# Expected value

(Redirected from Expected number)

In probability theory, the expected value of a random variable, intuitively, is the long-run average value of repetitions of the same experiment it represents. For example, the expected value in rolling a six-sided die is 3.5, because the average of all the numbers that come up is 3.5 as the number of rolls approaches infinity (see § Examples for details). In other words, the law of large numbers states that the arithmetic mean of the values almost surely converges to the expected value as the number of repetitions approaches infinity. The expected value is also known as the expectation, mathematical expectation, EV, average, mean value, mean, or first moment.

More practically, the expected value of a discrete random variable is the probability-weighted average of all possible values. In other words, each possible value the random variable can assume is multiplied by its probability of occurring, and the resulting products are summed to produce the expected value. The same principle applies to an absolutely continuous random variable, except that an integral of the variable with respect to its probability density replaces the sum. The formal definition subsumes both of these and also works for distributions which are neither discrete nor absolutely continuous; the expected value of a random variable is the integral of the random variable with respect to its probability measure.[1][2]

The expected value does not exist for random variables having some distributions with large "tails", such as the Cauchy distribution.[3] For random variables such as these, the long-tails of the distribution prevent the sum or integral from converging.

The expected value is a key aspect of how one characterizes a probability distribution; it is one type of location parameter. By contrast, the variance is a measure of dispersion of the possible values of the random variable around the expected value. The variance itself is defined in terms of two expectations: it is the expected value of the squared deviation of the variable's value from the variable's expected value (var(X) = E[(X - E[X])2] = E(X2) - [E(X)]2).

The expected value plays important roles in a variety of contexts. In regression analysis, one desires a formula in terms of observed data that will give a "good" estimate of the parameter giving the effect of some explanatory variable upon a dependent variable. The formula will give different estimates using different samples of data, so the estimate it gives is itself a random variable. A formula is typically considered good in this context if it is an unbiased estimator— that is if the expected value of the estimate (the average value it would give over an arbitrarily large number of separate samples) can be shown to equal the true value of the desired parameter.

In decision theory, and in particular in choice under uncertainty, an agent is described as making an optimal choice in the context of incomplete information. For risk neutral agents, the choice involves using the expected values of uncertain quantities, while for risk averse agents it involves maximizing the expected value of some objective function such as a von Neumann–Morgenstern utility function. One example of using expected value in reaching optimal decisions is the Gordon–Loeb model of information security investment. According to the model, one can conclude that the amount a firm spends to protect information should generally be only a small fraction of the expected loss (i.e., the expected value of the loss resulting from a cyber or information security breach).[4]

## Definition

### Finite case

Let ${\displaystyle X}$  be a random variable with a finite number of finite outcomes ${\displaystyle x_{1}}$ , ${\displaystyle x_{2}}$ , ..., ${\displaystyle x_{k}}$  occurring with probabilities ${\displaystyle p_{1}}$ , ${\displaystyle p_{2}}$ , ..., ${\displaystyle p_{k}}$ , respectively. The expectation of ${\displaystyle X}$  is defined as

${\displaystyle \operatorname {E} [X]=\sum _{i=1}^{k}x_{i}\,p_{i}=x_{1}p_{1}+x_{2}p_{2}+\cdots +x_{k}p_{k}.}$

Since all probabilities ${\displaystyle p_{i}}$  add up to 1 (${\displaystyle p_{1}+p_{2}+\cdots +p_{k}=1}$ ), the expected value is the weighted average, with ${\displaystyle p_{i}}$ ’s being the weights.

If all outcomes ${\displaystyle x_{i}}$  are equiprobable (that is, ${\displaystyle p_{1}=p_{2}=\cdots =p_{k}}$ ), then the weighted average turns into the simple average. This is intuitive: the expected value of a random variable is the average of all values it can take; thus the expected value is what one expects to happen on average. If the outcomes ${\displaystyle x_{i}}$  are not equiprobable, then the simple average must be replaced with the weighted average, which takes into account the fact that some outcomes are more likely than the others. The intuition however remains the same: the expected value of ${\displaystyle X}$  is what one expects to happen on average.

An illustration of the convergence of sequence averages of rolls of a die to the expected value of 3.5 as the number of rolls (trials) grows.

#### Examples

• Let ${\displaystyle X}$  represent the outcome of a roll of a fair six-sided die. More specifically, ${\displaystyle X}$  will be the number of pips showing on the top face of the die after the toss. The possible values for ${\displaystyle X}$  are 1, 2, 3, 4, 5, and 6, all of which are equally likely with a probability of 1/6. The expectation of ${\displaystyle X}$  is
${\displaystyle \operatorname {E} [X]=1\cdot {\frac {1}{6}}+2\cdot {\frac {1}{6}}+3\cdot {\frac {1}{6}}+4\cdot {\frac {1}{6}}+5\cdot {\frac {1}{6}}+6\cdot {\frac {1}{6}}=3.5.}$
If one rolls the die ${\displaystyle n}$  times and computes the average (arithmetic mean) of the results, then as ${\displaystyle n}$  grows, the average will almost surely converge to the expected value, a fact known as the strong law of large numbers. One example sequence of ten rolls of the die is 2, 3, 1, 2, 5, 6, 2, 2, 2, 6, which has the average of 3.1, with the distance of 0.4 from the expected value of 3.5. The convergence is relatively slow: the probability that the average falls within the range 3.5 ± 0.1 is 21.6% for ten rolls, 46.1% for a hundred rolls and 93.7% for a thousand rolls. See the figure for an illustration of the averages of longer sequences of rolls of the die and how they converge to the expected value of 3.5. More generally, the rate of convergence can be roughly quantified by e.g. Chebyshev's inequality and the Berry–Esseen theorem.
• The roulette game consists of a small ball and a wheel with 38 numbered pockets around the edge. As the wheel is spun, the ball bounces around randomly until it settles down in one of the pockets. Suppose random variable ${\displaystyle X}$  represents the (monetary) outcome of a $1 bet on a single number ("straight up" bet). If the bet wins (which happens with probability 1/38 in American roulette), the payoff is$35; otherwise the player loses the bet. The expected profit from such a bet will be
${\displaystyle \operatorname {E} [\,{\text{gain from }}\1{\text{ bet}}\,]=-\1\cdot {\frac {37}{38}}+\35\cdot {\frac {1}{38}}=-\0.0526.}$
That is, the bet of $1 stands to lose$0.0526, so its expected value is -\$0.0526.

### Countably infinite case

Let ${\displaystyle X}$  be a random variable with a countable set of finite outcomes ${\displaystyle x_{1}}$ , ${\displaystyle x_{2}}$ , ..., occurring with probabilities ${\displaystyle p_{1}}$ , ${\displaystyle p_{2}}$ , ..., respectively, such that the infinite sum ${\displaystyle \textstyle \sum _{i=1}^{\infty }|x_{i}|\,p_{i}}$  converges. The expected value of ${\displaystyle X}$  is defined as the series

${\displaystyle \operatorname {E} [X]=\sum _{i=1}^{\infty }x_{i}\,p_{i}.}$

Remark 1. Observe that ${\displaystyle \textstyle {\Bigl |}\operatorname {E} [X]{\Bigr |}\leq \sum _{i=1}^{\infty }|x_{i}|\,p_{i}<\infty .}$

Remark 2. Due to absolute convergence, the expected value does not depend on the order in which the outcomes are presented. By contrast, a conditionally convergent series can be made to converge or diverge arbitrarily, via the Riemann rearrangement theorem.

#### Example

• Suppose ${\displaystyle x_{i}=i}$  and ${\displaystyle p_{i}={\frac {k}{i2^{i}}},}$  for ${\displaystyle i=1,2,3,\ldots }$ , where ${\displaystyle k={\frac {1}{\ln 2}}}$  (with ${\displaystyle \ln }$  being the natural logarithm) is the scale factor such that the probabilities sum to 1. Then
${\displaystyle \operatorname {E} [X]=1\left({\frac {k}{2}}\right)+2\left({\frac {k}{8}}\right)+3\left({\frac {k}{24}}\right)+\dots ={\frac {k}{2}}+{\frac {k}{4}}+{\frac {k}{8}}+\dots =k.}$
Since this series converges absolutely, the expected value of ${\displaystyle X}$  is ${\displaystyle k}$ .
• For an example that is not absolutely convergent, suppose random variable ${\displaystyle X}$  takes values 1, −2, 3, −4, ..., with respective probabilities ${\displaystyle {\frac {c}{1^{2}}},{\frac {c}{2^{2}}},{\frac {c}{3^{2}}},{\frac {c}{4^{2}}}}$ , ..., where ${\displaystyle c={\frac {6}{\pi ^{2}}}}$  is a normalizing constant that ensures the probabilities sum up to one. Then the infinite sum
${\displaystyle \sum _{i=1}^{\infty }x_{i}\,p_{i}=c\,{\bigg (}1-{\frac {1}{2}}+{\frac {1}{3}}-{\frac {1}{4}}+\dotsb {\bigg )}}$
converges and its sum is equal to ${\displaystyle {\frac {6\ln 2}{\pi ^{2}}}\approx 0.421383}$ . However it would be incorrect to claim that the expected value of ${\displaystyle X}$  is equal to this number—in fact ${\displaystyle \operatorname {E} [X]}$  does not exist (finite or infinite), as this series does not converge absolutely (see Alternating harmonic series).
• An example that diverges arises in the context of the St. Petersburg paradox. Let ${\displaystyle x_{i}=2^{i}}$  and ${\displaystyle p_{i}={\frac {1}{2^{i}}}}$  for ${\displaystyle i=1,2,3,\ldots }$ . The expected value calculation gives
${\displaystyle \sum _{i=1}^{\infty }x_{i}\,p_{i}=2\cdot {\frac {1}{2}}+4\cdot {\frac {1}{4}}+8\cdot {\frac {1}{8}}+16\cdot {\frac {1}{16}}+\cdots =1+1+1+1+\cdots \,.}$
Since this does not converge but instead keeps growing, the expected value is infinite.

### Absolutely continuous case

If ${\displaystyle X}$  is a random variable whose cumulative distribution function admits a density ${\displaystyle f(x)}$ , then the expected value is defined as the following Lebesgue integral:

${\displaystyle \operatorname {E} [X]=\int _{\mathbb {R} }xf(x)\,dx.}$

Remark. From computational perspective, the integral in the definition of ${\displaystyle \operatorname {E} [X]}$  may often be treated as an improper Riemann integral ${\displaystyle \textstyle \int _{-\infty }^{+\infty }xf(x)\,dx.}$  Specifically, if the function ${\displaystyle xf(x)}$  is Riemann-integrable on every finite interval ${\displaystyle [a,b]}$ , and

${\displaystyle \min \left((-1)\cdot {\hbox{(R)}}\int _{-\infty }^{0}xf(x)\,dx,\ {\hbox{(R)}}\int _{0}^{+\infty }xf(x)\,dx\right)<\infty ,}$

then the values (whether finite or infinite) of both integrals agree.

### General case

In general, if ${\displaystyle X}$  is a random variable defined on a probability space ${\displaystyle (\Omega ,\Sigma ,\operatorname {P} )}$ , then the expected value of ${\displaystyle X}$ , denoted by ${\displaystyle \operatorname {E} [X]}$ , ${\displaystyle \langle X\rangle }$ , or ${\displaystyle {\bar {X}}}$ , is defined as the Lebesgue integral

${\displaystyle \operatorname {E} [X]=\int _{\Omega }X(\omega )\,d\operatorname {P} (\omega ).}$

Remark 1. If ${\displaystyle X_{+}(\omega )=\max(X(\omega ),0)}$  and ${\displaystyle X_{-}(\omega )=-\min(X(\omega ),0)}$ , then ${\displaystyle X=X_{+}-X_{-}.}$  The functions ${\displaystyle X_{+}}$  and ${\displaystyle X_{-}}$  can be shown to be measurable (hence, random variables), and, by definition of Lebesgue integral,

{\displaystyle {\begin{aligned}\operatorname {E} [X]&=\int _{\Omega }X(\omega )\,d\operatorname {P} (\omega )\\&=\int _{\Omega }X_{+}(\omega )\,d\operatorname {P} (\omega )-\int _{\Omega }X_{-}(\omega )\,d\operatorname {P} (\omega )\\&=\operatorname {E} [X_{+}]-\operatorname {E} [X_{-}],\end{aligned}}}

where ${\displaystyle \operatorname {E} [X_{+}]}$  and ${\displaystyle \operatorname {E} [X_{-}]}$  are non-negative and possibly infinite.

The following scenarios are possible:

• ${\displaystyle \operatorname {E} [X]}$  is finite, i.e. ${\displaystyle \max(\operatorname {E} [X_{+}],\operatorname {E} [X_{-}])<\infty ;}$
• ${\displaystyle \operatorname {E} [X]}$  is infinite, i.e. ${\displaystyle \max(\operatorname {E} [X_{+}],\operatorname {E} [X_{-}])=\infty }$  and ${\displaystyle \min(\operatorname {E} [X_{+}],\operatorname {E} [X_{-}])<\infty ;}$
• ${\displaystyle \operatorname {E} [X]}$  is neither finite nor infinite, i.e. ${\displaystyle \operatorname {E} [X_{+}]=\operatorname {E} [X_{-}]=\infty .}$

Remark 2. If ${\displaystyle F_{X}(x)=\operatorname {P} (X\leq x)}$  is the cumulative distribution function of ${\displaystyle X}$ , then

${\displaystyle \operatorname {E} [X]=\int _{-\infty }^{+\infty }x\,dF_{X}(x),}$

where the integral is interpreted in the sense of Lebesgue–Stieltjes.

Remark 3. An example of a distribution for which there is no expected value is Cauchy distribution.

Remark 4. For multidimensional random variables, their expected value is defined per component, i.e.

${\displaystyle \operatorname {E} [(X_{1},\ldots ,X_{n})]=(\operatorname {E} [X_{1}],\ldots ,\operatorname {E} [X_{n}])}$

and, for a random matrix ${\displaystyle X}$  with elements ${\displaystyle X_{ij}}$ ,

${\displaystyle (\operatorname {E} [X])_{ij}=\operatorname {E} [X_{ij}].}$

## Basic properties

The properties below replicate or follow immediately from those of Lebesgue integral.

### ${\displaystyle \operatorname {E} [{\mathbf {1} }_{A}]=\operatorname {P} (A)}$

If ${\displaystyle A}$  is an event, then ${\displaystyle \operatorname {E} [{\mathbf {1} }_{A}]=\operatorname {P} (A),}$  where ${\displaystyle {\mathbf {1} }_{A}}$  is the indicator function of the set ${\displaystyle A}$ .

Proof. By definition of Lebesgue integral of the simple function ${\displaystyle {\mathbf {1} }_{A}={\mathbf {1} }_{A}(\omega )}$ ,

${\displaystyle \operatorname {E} [{\mathbf {1} }_{A}]=1\cdot \operatorname {P} (A)+0\cdot \operatorname {P} (\Omega \setminus A)=\operatorname {P} (A).}$

### If X = Y (a.s.) then E[X] = E[Y]

The statement follows from the definition of Lebesgue integral if we notice that ${\displaystyle X_{+}=Y_{+}}$  (a.s.), ${\displaystyle X_{-}=Y_{-}}$  (a.s.), and that changing a simple random variable on a set of probability zero does not alter the expected value.

### Expected value of a constant

If ${\displaystyle X}$  is a random variable, and ${\displaystyle X=c}$  (a.s.), where ${\displaystyle c\in [-\infty ,+\infty ]}$ , then ${\displaystyle \operatorname {E} [X]=c}$ . In particular, for an arbitrary random variable ${\displaystyle X}$ , ${\displaystyle \operatorname {E} [\operatorname {E} [X]]=\operatorname {E} [X]}$ .

### Linearity

The expected value operator (or expectation operator) ${\displaystyle \operatorname {E} [\cdot ]}$  is linear in the sense that

{\displaystyle {\begin{aligned}\operatorname {E} [X+Y]&=\operatorname {E} [X]+\operatorname {E} [Y],\\[6pt]\operatorname {E} [aX]&=a\operatorname {E} [X],\end{aligned}}}

where ${\displaystyle X}$  and ${\displaystyle Y}$  are arbitrary random variables, and ${\displaystyle a}$  is a constant.

More rigorously, let ${\displaystyle X}$  and ${\displaystyle Y}$  be random variables whose expected values are defined (different from ${\displaystyle \infty -\infty }$ ).

• If ${\displaystyle \operatorname {E} [X]+\operatorname {E} [Y]}$  is also defined (i.e. differs from ${\displaystyle \infty -\infty }$ ), then
${\displaystyle \operatorname {E} [X+Y]=\operatorname {E} [X]+\operatorname {E} [Y].}$
• Let ${\displaystyle \operatorname {E} [X]}$  be finite, and ${\displaystyle a\in \mathbb {R} }$  be a finite scalar. Then ${\displaystyle \operatorname {E} [aX]=a\operatorname {E} [X].}$

### E[X] exists and is finite if and only if E[|X|] is finite

The following statements regarding a random variable ${\displaystyle X}$  are equivalent:

• ${\displaystyle \operatorname {E} [X]}$  exists and is finite.
• Both ${\displaystyle \operatorname {E} [X_{+}]}$  and ${\displaystyle \operatorname {E} [X_{-}]}$  are finite.
• ${\displaystyle \operatorname {E} [|X|]}$  is finite.

Sketch of proof. Indeed, ${\displaystyle |X|=X_{+}+X_{-}}$ . By linearity, ${\displaystyle \operatorname {E} [|X|]=\operatorname {E} [X_{+}]+\operatorname {E} [X_{-}]}$ . The above equivalency relies on the definition of Lebesgue integral and measurability of ${\displaystyle X}$ .

Remark. For the reasons above, the expressions "${\displaystyle X}$  is integrable" and "the expected value of ${\displaystyle X}$  is finite" are used interchangeably when speaking of a random variable throughout this article.

### Monotonicity

If ${\displaystyle X\leq Y}$  (a.s.), and both ${\displaystyle \operatorname {E} [X]}$  and ${\displaystyle \operatorname {E} [Y]}$  exist, then ${\displaystyle \operatorname {E} [X]\leq \operatorname {E} [Y]}$ .

Remark. ${\displaystyle \operatorname {E} [X]}$  and ${\displaystyle \operatorname {E} [Y]}$  exist in the sense that ${\displaystyle \min(\operatorname {E} [X_{+}],\operatorname {E} [X_{-}])<\infty }$  and ${\displaystyle \min(\operatorname {E} [Y_{+}],\operatorname {E} [Y_{-}])<\infty .}$

Proof follows from the linearity and the previous property if we set ${\displaystyle Z=Y-X}$  and notice that ${\displaystyle Z\geq 0}$  (a.s.).

### If ${\displaystyle |X|\leq Y}$  (a.s.) and ${\displaystyle \operatorname {E} [Y]}$  is finite then so is ${\displaystyle \operatorname {E} [X]}$

Let ${\displaystyle X}$  and ${\displaystyle Y}$  be random variables such that ${\displaystyle |X|\leq Y}$  (a.s.) and ${\displaystyle \operatorname {E} [Y]<\infty }$ . Then ${\displaystyle \operatorname {E} [X]\neq \pm \infty }$ .

Proof. Due to non-negativity of ${\displaystyle |X|}$ , ${\displaystyle \operatorname {E} |X|}$  exists, finite or infinite. By monotonicity, ${\displaystyle \operatorname {E} |X|\leq \operatorname {E} [Y]<\infty }$ , so ${\displaystyle \operatorname {E} |X|}$  is finite which, as we saw earlier, is equivalent to ${\displaystyle \operatorname {E} [X]}$  being finite.

### If ${\displaystyle \operatorname {E} |X^{\beta }|<\infty }$  and ${\displaystyle 0<\alpha <\beta }$  then ${\displaystyle \operatorname {E} |X^{\alpha }|<\infty }$

The proposition below will be used to prove the extremal property of ${\displaystyle \operatorname {E} [X]}$  later on.

Proposition. If ${\displaystyle X}$  is a random variable, then so is ${\displaystyle X^{\alpha }}$ , for every ${\displaystyle \alpha >0}$ . If, in addition, ${\displaystyle \operatorname {E} |X^{\beta }|<\infty }$  and ${\displaystyle 0<\alpha <\beta }$ , then ${\displaystyle \operatorname {E} |X^{\alpha }|<\infty }$ .

#### Counterexample for infinite measure

The requirement that ${\displaystyle \operatorname {P} (\Omega )<\infty }$  is essential. By way of counterexample, consider the measurable space

${\displaystyle ([1,+\infty ),{\mathcal {B}}_{\mathbb {R} _{[1,+\infty )}},\lambda ),}$

where ${\displaystyle {\mathcal {B}}_{\mathbb {R} _{[1,+\infty )}}}$  is the Borel ${\displaystyle \sigma }$ -algebra on the interval ${\displaystyle [1,+\infty ),}$  and ${\displaystyle \lambda }$  is the linear Lebesgue measure. The reader can prove that ${\displaystyle \textstyle \int _{[1,+\infty )}{\frac {1}{x}}\,dx=\infty ,}$  even though ${\displaystyle \textstyle \int _{[1,+\infty )}{\frac {1}{x^{2}}}\,dx=1.}$  (Sketch of proof: ${\displaystyle \textstyle \int _{S}{\frac {1}{x}}\,dx}$  and ${\displaystyle \textstyle \int _{S}{\frac {1}{x^{2}}}\,dx}$  define a measure ${\displaystyle \mu }$  on ${\displaystyle \textstyle [1,+\infty )=\cup _{n=1}^{\infty }[1,n].}$  Use "continuity from below" w.r. to ${\displaystyle \mu }$  and reduce to Riemann integral on each finite subinterval ${\displaystyle [1,n]}$ ).

### Extremal property

Recall, as we proved early on, that if ${\displaystyle X}$  is a random variable, then so is ${\displaystyle X^{2}}$ .

Proposition (extremal property of ${\displaystyle \operatorname {E} [X])}$ ). Let ${\displaystyle X}$  be a random variable, and ${\displaystyle \operatorname {E} [X^{2}]<\infty }$ . Then ${\displaystyle \operatorname {E} [X]}$  and ${\displaystyle \operatorname {Var} [X]}$  are finite, and ${\displaystyle \operatorname {E} [X]}$  is the best least squares approximation for ${\displaystyle X}$  among constants. Specifically,

• for every ${\displaystyle c\in \mathbb {R} }$ , ${\displaystyle \textstyle \operatorname {E} [X-c]^{2}\geq \operatorname {Var} [X];}$
• equality holds if and only if ${\displaystyle c=\operatorname {E} [X].}$

(${\displaystyle \operatorname {Var} [X]}$  denotes the variance of ${\displaystyle X}$ ).

Remark (intuitive interpretation of extremal property). In intuitive terms, the extremal property says that if one is asked to predict the outcome of a trial of a random variable ${\displaystyle X}$ , then ${\displaystyle \operatorname {E} [X]}$ , in some practically useful sense, is one's best bet if no advance information about the outcome is available. If, on the other hand, one does have some advance knowledge ${\displaystyle {\mathcal {F}}}$  regarding the outcome, then — again, in some practically useful sense — one's bet may be improved upon by using conditional expectations ${\displaystyle \operatorname {E} [X\mid {\mathcal {F}}]}$  (of which ${\displaystyle \operatorname {E} [X]}$  is a special case) rather than ${\displaystyle \operatorname {E} [X]}$ .

Proof of proposition. By the above properties, both ${\displaystyle \operatorname {E} [X]}$  and ${\displaystyle \operatorname {Var} [X]=\operatorname {E} [X^{2}]-\operatorname {E} ^{2}[X]}$  are finite, and

{\displaystyle {\begin{aligned}\operatorname {E} [X-c]^{2}&=\operatorname {E} [X^{2}-2cX+c^{2}]\\[6pt]&=\operatorname {E} [X^{2}]-2c\operatorname {E} [X]+c^{2}\\[6pt]&=(c-\operatorname {E} [X])^{2}+\operatorname {E} [X^{2}]-\operatorname {E} ^{2}[X]\\[6pt]&=(c-\operatorname {E} [X])^{2}+\operatorname {Var} [X],\end{aligned}}}

whence the extremal property follows.

### Non-degeneracy

If ${\displaystyle \operatorname {E} |X|=0}$ , then ${\displaystyle X=0}$  (a.s.).

### ${\displaystyle |\operatorname {E} [X]|\leq \operatorname {E} |X|}$

For an arbitrary random variable ${\displaystyle X}$ , ${\displaystyle |\operatorname {E} [X]|\leq \operatorname {E} |X|}$ .

Proof. By definition of Lebesgue integral,

{\displaystyle {\begin{aligned}|\operatorname {E} [X]|&={\Bigl |}\operatorname {E} [X_{+}]-\operatorname {E} [X_{-}]{\Bigr |}\leq {\Bigl |}\operatorname {E} [X_{+}]{\Bigr |}+{\Bigl |}\operatorname {E} [X_{-}]{\Bigr |}\\[5pt]&=\operatorname {E} [X_{+}]+\operatorname {E} [X_{-}]=\operatorname {E} [X_{+}+X_{-}]\\[5pt]&=\operatorname {E} |X|.\end{aligned}}}

Note that this result can also be proved based on Jensen's inequality.

### Non-multiplicativity

In general, the expected value operator is not multiplicative, i.e. ${\displaystyle \operatorname {E} [XY]}$  is not necessarily equal to ${\displaystyle \operatorname {E} [X]\cdot \operatorname {E} [Y]}$ . Indeed, let ${\displaystyle X}$  assume the values of 1 and -1 with probability 0.5 each. Then

${\displaystyle \operatorname {E^{2}} [X]=\left({\frac {1}{2}}\cdot (-1)+{\frac {1}{2}}\cdot 1\right)^{2}=0,}$

and

${\displaystyle \operatorname {E} [X^{2}]={\frac {1}{2}}\cdot (-1)^{2}+{\frac {1}{2}}\cdot 1^{2}=1,{\text{ so }}\operatorname {E} [X^{2}]\neq \operatorname {E^{2}} [X].}$

The amount by which the multiplicativity fails is called the covariance:

${\displaystyle \operatorname {Cov} (X,Y)=\operatorname {E} [XY]-\operatorname {E} [X]\operatorname {E} [Y].}$

However, if ${\displaystyle X}$  and ${\displaystyle Y}$  are independent, then ${\displaystyle \operatorname {E} [XY]=\operatorname {E} [X]\operatorname {E} [Y]}$ , and ${\displaystyle \operatorname {Cov} (X,Y)=0}$ .

### Counterexample: ${\displaystyle \operatorname {E} [X_{i}]\not \to \operatorname {E} [X]}$  despite ${\displaystyle X_{i}\to X}$  pointwise

Let ${\displaystyle \left([0,1],{\mathcal {B}}_{[0,1]},{\mathrm {P} }\right)}$  be the probability space, where ${\displaystyle {\mathcal {B}}_{[0,1]}}$  is the Borel ${\displaystyle \sigma }$ -algebra on ${\displaystyle [0,1]}$  and ${\displaystyle {\mathrm {P} }}$  the linear Lebesgue measure. For ${\displaystyle i\geq 1,}$  define a sequence of random variables

${\displaystyle X_{i}=i\cdot {\mathbf {1} }_{\left[0,{\frac {1}{i}}\right]}}$

and a random variable

${\displaystyle X={\begin{cases}+\infty &{\text{if}}\ x=0\\0&{\text{otherwise.}}\end{cases}}}$

on ${\displaystyle [0,1]}$ , with ${\displaystyle {\mathbf {1} }_{S}}$  being the indicator function of the set ${\displaystyle S\subseteq [0,1]}$ .

For every ${\displaystyle x\in [0,1],}$  as ${\displaystyle i\to +\infty ,}$  ${\displaystyle X_{i}(x)\to X(x),}$  and

${\displaystyle \operatorname {E} [X_{i}]=i\cdot {\mathrm {P} }\left(\left[0,{\frac {1}{i}}\right]\right)=i\cdot {\dfrac {1}{i}}=1,}$

so ${\displaystyle \lim _{i\to \infty }\operatorname {E} [X_{i}]=1.}$  On the other hand, ${\displaystyle \mathop {\mathrm {P} } (\{0\})=0,}$  and hence ${\displaystyle \operatorname {E} \left[X\right]=0.}$

In general, the expected value operator is not ${\displaystyle \sigma }$ -additive, i.e.

${\displaystyle \operatorname {E} \left[\sum _{i=0}^{\infty }X_{i}\right]\neq \sum _{i=0}^{\infty }\operatorname {E} [X_{i}].}$

By way of counterexample, let ${\displaystyle \left([0,1],{\mathcal {B}}_{[0,1]},{\mathrm {P} }\right)}$  be the probability space, where ${\displaystyle {\mathcal {B}}_{[0,1]}}$  is the Borel ${\displaystyle \sigma }$ -algebra on ${\displaystyle [0,1]}$  and ${\displaystyle {\mathrm {P} }}$  the linear Lebesgue measure. Define a sequence of random variables ${\displaystyle \textstyle X_{i}=(i+1)\cdot {\mathbf {1} }_{\left[0,{\frac {1}{i+1}}\right]}-i\cdot {\mathbf {1} }_{\left[0,{\frac {1}{i}}\right]}}$  on ${\displaystyle [0,1]}$ , with ${\displaystyle {\mathbf {1} }_{S}}$  being the indicator function of the set ${\displaystyle S\subseteq [0,1]}$ . For the pointwise sums, we have

${\displaystyle \sum _{i=0}^{n}X_{i}=(n+1)\cdot {\mathbf {1} }_{\left[0,{\frac {1}{n+1}}\right]},}$
${\displaystyle \sum _{i=0}^{\infty }X_{i}(x)={\begin{cases}+\infty &{\text{if}}\ x=0\\0&{\text{otherwise.}}\end{cases}}}$

${\displaystyle \sum _{i=0}^{\infty }\operatorname {E} [X_{i}]=\lim _{n\to \infty }\sum _{i=0}^{n}\operatorname {E} [X_{i}]=\lim _{n\to \infty }\operatorname {E} \left[\sum _{i=0}^{n}X_{i}\right]=1.}$

On the other hand, ${\displaystyle \mathop {\mathrm {P} } (\{0\})=0,}$  and hence

${\displaystyle \operatorname {E} \left[\sum _{i=0}^{\infty }X_{i}\right]=0\neq 1=\sum _{i=0}^{\infty }\operatorname {E} [X_{i}].}$

### Countable additivity for non-negative random variables

Let ${\displaystyle \{X_{i}\}_{i=0}^{\infty }}$  be non-negative random variables. It follows from monotone convergence theorem that

${\displaystyle \operatorname {E} \left[\sum _{i=0}^{\infty }X_{i}\right]=\sum _{i=0}^{\infty }\operatorname {E} [X_{i}].}$

## Inequalities

### Cauchy–Bunyakovsky–Schwarz inequality

The Cauchy–Bunyakovsky–Schwarz inequality states that

${\displaystyle (\operatorname {E} [XY])^{2}\leq \operatorname {E} [X^{2}]\cdot \operatorname {E} [Y^{2}].}$

### Markov's inequality

For a nonnegative random variable ${\displaystyle X}$  and ${\displaystyle a>0}$ , Markov's inequality states that

${\displaystyle \operatorname {P} (X\geq a)\leq {\frac {\operatorname {E} [X]}{a}}.}$

### Bienaymé-Chebyshev inequality

Let ${\displaystyle X}$  be an arbitrary random variable with finite expected value ${\displaystyle \operatorname {E} [X]}$  and finite variance ${\displaystyle \operatorname {Var} [X]\neq 0}$ . The Bienaymé-Chebyshev inequality states that, for any real number ${\displaystyle k>0}$ ,

${\displaystyle \operatorname {P} {\Bigl (}{\Bigl |}X-\operatorname {E} [X]{\Bigr |}\geq k{\sqrt {\operatorname {Var} [X]}}{\Bigr )}\leq {\frac {1}{k^{2}}}.}$

### Jensen's inequality

Let ${\displaystyle f:{\mathbb {R} }\to {\mathbb {R} }}$  be a Borel convex function and ${\displaystyle X}$  a random variable such that ${\displaystyle \operatorname {E} |X|<\infty }$ . Jensen's inequality states that

${\displaystyle f(\operatorname {E} (X))\leq \operatorname {E} (f(X)).}$

Remark 1. The expected value ${\displaystyle \operatorname {E} (f(X))}$  is well-defined even if ${\displaystyle X}$  is allowed to assume infinite values. Indeed, ${\displaystyle \operatorname {E} |X|<\infty }$  implies that ${\displaystyle X\neq \pm \infty }$  (a.s.), so the random variable ${\displaystyle f(X(\omega ))}$  is defined almost sure, and therefore there is enough information to compute ${\displaystyle \operatorname {E} (f(X)).}$

Remark 2. Jensen's inequality implies that ${\displaystyle |\operatorname {E} [X]|\leq \operatorname {E} |X|}$  since the absolute value function is convex.

### Lyapunov's inequality

Let ${\displaystyle 0 . Lyapunov's inequality states that

${\displaystyle {\Bigl (}\operatorname {E} |X|^{s}{\Bigr )}^{1/s}\leq \left(\operatorname {E} |X|^{t}\right)^{1/t}.}$

Proof. Applying Jensen's inequality to ${\displaystyle |X|^{s}}$  and ${\displaystyle g(x)=|x|^{t/s}}$ , obtain ${\displaystyle {\Bigl |}\operatorname {E} |X^{s}|{\Bigr |}^{t/s}\leq \operatorname {E} |X^{s}|^{t/s}=\operatorname {E} |X|^{t}}$ . Taking the ${\displaystyle t}$ th root of each side completes the proof.

Corollary.

${\displaystyle \operatorname {E} |X|\leq {\Bigl (}\operatorname {E} |X|^{2}{\Bigr )}^{1/2}\leq \cdots \leq {\Bigl (}\operatorname {E} |X|^{n}{\Bigr )}^{1/n}\leq \cdots }$

### Hölder's inequality

Let ${\displaystyle p}$  and ${\displaystyle q}$  satisfy ${\displaystyle 1\leq p\leq \infty }$ , ${\displaystyle 1\leq q\leq \infty }$ , and ${\displaystyle 1/p+1/q=1}$ . The Hölder's inequality states that

${\displaystyle \operatorname {E} |XY|\leq (\operatorname {E} |X|^{p})^{1/p}(\operatorname {E} |Y|^{q})^{1/q}.}$

### Minkowski inequality

Let ${\displaystyle p}$  be an integer satisfying ${\displaystyle 1\leq p\leq \infty }$ . Let, in addition, ${\displaystyle \operatorname {E} |X|^{p}<\infty }$  and ${\displaystyle \operatorname {E} |Y|^{p}<\infty }$ . Then, according to the Minkowski inequality, ${\displaystyle \operatorname {E} |X+Y|^{p}<\infty }$  and

${\displaystyle {\Bigl (}\operatorname {E} |X+Y|^{p}{\Bigr )}^{1/p}\leq {\Bigl (}\operatorname {E} |X|^{p}{\Bigr )}^{1/p}+{\Bigl (}\operatorname {E} |Y|^{p}{\Bigr )}^{1/p}.}$

## Taking limits under the ${\displaystyle \operatorname {E} }$ sign

### Monotone convergence theorem

Let the sequence of random variables ${\displaystyle \{X_{n}\}}$  and the random variables ${\displaystyle X}$  and ${\displaystyle Y}$  be defined on the same probability space ${\displaystyle (\Omega ,\Sigma ,\operatorname {P} ).}$  Suppose that

• all the expected values ${\displaystyle \operatorname {E} [X_{n}],}$  ${\displaystyle \operatorname {E} [X],}$  and ${\displaystyle \operatorname {E} [Y]}$  are defined (differ from ${\displaystyle \infty -\infty }$ );
• ${\displaystyle \operatorname {E} [Y]>-\infty ;}$
• for every ${\displaystyle n,}$
${\displaystyle -\infty \leq Y\leq X_{n}\leq X_{n+1}\leq +\infty \quad {\hbox{(a.s.)}};}$
• ${\displaystyle X}$  is the pointwise limit of ${\displaystyle \{X_{n}\}}$  (a.s.), i.e. ${\displaystyle X(\omega )=\lim \nolimits _{n}X_{n}(\omega )}$  (a.s.).

The monotone convergence theorem states that

${\displaystyle \lim _{n}\operatorname {E} [X_{n}]=\operatorname {E} [X].}$

### Fatou's lemma

Let the sequence of random variables ${\displaystyle \{X_{n}\}}$  and the random variable ${\displaystyle Y}$  be defined on the same probability space ${\displaystyle (\Omega ,\Sigma ,\operatorname {P} ).}$  Suppose that

• all the expected values ${\displaystyle \operatorname {E} [X_{n}],}$  ${\displaystyle \textstyle \operatorname {E} [\liminf _{n}X_{n}],}$  and ${\displaystyle \operatorname {E} [Y]}$  are defined (differ from ${\displaystyle \infty -\infty }$ );
• ${\displaystyle \operatorname {E} [Y]>-\infty ;}$
• ${\displaystyle -\infty \leq Y\leq X_{n}\leq +\infty }$  (a.s.), for every ${\displaystyle n.}$

Fatou's lemma states that

${\displaystyle \operatorname {E} [\liminf _{n}X_{n}]\leq \liminf _{n}\operatorname {E} [X_{n}].}$

(Note that ${\displaystyle \textstyle \liminf _{n}X_{n}}$  is a random variable, for every ${\displaystyle n,}$  by the properties of limit inferior).

Corollary. Let

• ${\displaystyle X_{n}\to X}$  pointwise (a.s.);
• ${\displaystyle \operatorname {E} [X_{n}]\leq C,}$  for some constant ${\displaystyle C}$  (independent from ${\displaystyle n}$ );
• ${\displaystyle \operatorname {E} [Y]>-\infty ;}$
• ${\displaystyle -\infty \leq Y\leq X_{n}\leq +\infty }$  (a.s.), for every ${\displaystyle n.}$

Then ${\displaystyle \operatorname {E} [X]\leq C.}$

Proof is by observing that ${\displaystyle \textstyle X=\liminf _{n}X_{n}}$  (a.s.) and applying Fatou's lemma.

### Dominated convergence theorem

Let ${\displaystyle \{X_{n}\}_{n}}$  be a sequence of random variables. If ${\displaystyle X_{n}\to X}$  pointwise (a.s.), ${\displaystyle |X_{n}|\leq Y\leq +\infty }$  (a.s.), and ${\displaystyle \operatorname {E} [Y]<\infty }$ . Then, according to the dominated convergence theorem,

• the function ${\displaystyle X}$  is measurable (hence a random variable);
• ${\displaystyle \operatorname {E} |X|<\infty }$ ;
• all the expected values ${\displaystyle \operatorname {E} [X_{n}]}$  and ${\displaystyle \operatorname {E} [X]}$  are defined (do not have the form ${\displaystyle \infty -\infty }$ );
• ${\displaystyle \lim _{n}\operatorname {E} [X_{n}]=\operatorname {E} [X]}$  (both sides may be infinite);
• ${\displaystyle \lim _{n}\operatorname {E} |X_{n}-X|=0.}$

### Uniform integrability

In some cases, the equality ${\displaystyle \displaystyle \lim _{n}\operatorname {E} [X_{n}]=\operatorname {E} [\lim _{n}X_{n}]}$  holds when the sequence ${\displaystyle \{X_{n}\}}$  is uniformly integrable.

## Relationship with characteristic function

The probability density function ${\displaystyle f_{X}}$  of a scalar random variable ${\displaystyle X}$  is related to its characteristic function ${\displaystyle \varphi _{X}}$  by the inversion formula:

${\displaystyle f_{X}(x)={\frac {1}{2\pi }}\int _{\mathbb {R} }e^{-itx}\varphi _{X}(t)\,dt.}$

For the expected value of ${\displaystyle g(X)}$  (where ${\displaystyle g:{\mathbb {R} }\to {\mathbb {R} }}$  is a Borel function), we can use this inversion formula to obtain

${\displaystyle \operatorname {E} [g(X)]={\frac {1}{2\pi }}\int _{\mathbb {R} }g(x)\left[\int _{\mathbb {R} }e^{-itx}\varphi _{X}(t)\,dt\right]\,dx.}$

If ${\displaystyle \operatorname {E} [g(X)]}$  is finite, changing the order of integration, we get, in accordance with Fubini–Tonelli theorem,

${\displaystyle \operatorname {E} [g(X)]={\frac {1}{2\pi }}\int _{\mathbb {R} }G(t)\varphi _{X}(t)\,dt,}$

where

${\displaystyle G(t)=\int _{\mathbb {R} }g(x)e^{-itx}\,dx}$

is the Fourier transform of ${\displaystyle g(x).}$  The expression for ${\displaystyle \operatorname {E} [g(X)]}$  also follows directly from Plancherel theorem.

## Uses and applications

It is possible to construct an expected value equal to the probability of an event by taking the expectation of an indicator function that is one if the event has occurred and zero otherwise. This relationship can be used to translate properties of expected values into properties of probabilities, e.g. using the law of large numbers to justify estimating probabilities by frequencies.

The expected values of the powers of X are called the moments of X; the moments about the mean of X are expected values of powers of X − E[X]. The moments of some random variables can be used to specify their distributions, via their moment generating functions.

To empirically estimate the expected value of a random variable, one repeatedly measures observations of the variable and computes the arithmetic mean of the results. If the expected value exists, this procedure estimates the true expected value in an unbiased manner and has the property of minimizing the sum of the squares of the residuals (the sum of the squared differences between the observations and the estimate). The law of large numbers demonstrates (under fairly mild conditions) that, as the size of the sample gets larger, the variance of this estimate gets smaller.

This property is often exploited in a wide variety of applications, including general problems of statistical estimation and machine learning, to estimate (probabilistic) quantities of interest via Monte Carlo methods, since most quantities of interest can be written in terms of expectation, e.g. ${\displaystyle \operatorname {P} ({X\in {\mathcal {A}}})=\operatorname {E} [{\mathbf {1} }_{\mathcal {A}}]}$ , where ${\displaystyle {\mathbf {1} }_{\mathcal {A}}}$  is the indicator function of the set ${\displaystyle {\mathcal {A}}}$ .

The mass of probability distribution is balanced at the expected value, here a Beta(α,β) distribution with expected value α/(α+β).

In classical mechanics, the center of mass is an analogous concept to expectation. For example, suppose X is a discrete random variable with values xi and corresponding probabilities pi. Now consider a weightless rod on which are placed weights, at locations xi along the rod and having masses pi (whose sum is one). The point at which the rod balances is E[X].

Expected values can also be used to compute the variance, by means of the computational formula for the variance

${\displaystyle \operatorname {Var} (X)=\operatorname {E} [X^{2}]-(\operatorname {E} [X])^{2}.}$

A very important application of the expectation value is in the field of quantum mechanics. The expectation value of a quantum mechanical operator ${\displaystyle {\hat {A}}}$  operating on a quantum state vector ${\displaystyle |\psi \rangle }$  is written as ${\displaystyle \langle {\hat {A}}\rangle =\langle \psi |A|\psi \rangle }$ . The uncertainty in ${\displaystyle {\hat {A}}}$  can be calculated using the formula ${\displaystyle (\Delta A)^{2}=\langle {\hat {A}}^{2}\rangle -\langle {\hat {A}}\rangle ^{2}}$ .

## The law of the unconscious statistician

The expected value of a measurable function of ${\displaystyle X}$ , ${\displaystyle g(X)}$ , given that ${\displaystyle X}$  has a probability density function ${\displaystyle f(x)}$ , is given by the inner product of ${\displaystyle f}$  and ${\displaystyle g}$ :

${\displaystyle \operatorname {E} [g(X)]=\int _{\mathbb {R} }g(x)f(x)\,dx.}$

This formula also holds in multidimensional case, when ${\displaystyle g}$  is a function of several random variables, and ${\displaystyle f}$  is their joint density.[5][6]

## Alternative formula for expected value

### Formula for non-negative random variables

#### Finite and countably infinite case

For a non-negative integer-valued random variable ${\displaystyle X:\Omega \to \{0,1,2,3,\ldots \}\cup \{+\infty \},}$

${\displaystyle \operatorname {E} [X]=\sum _{i=1}^{\infty }\operatorname {P} (X\geq i).}$