Expected Value of X Formula Continuous Random Variable
In the theory of probability, the expected value for any given random variable X is written as E (X), E [X]. It is a conception of the weighted arithmetic mean of a sizeable number of realizations of the random variable X that are independent. The alternate names for expectation are the mathematical expectation, average, expectation, mean, or first moment. Expected value is one of the key concepts in the fields of economics, finance and many more. The expectation of a constant is a constant. The expectation in different cases is discussed below.
a] Discrete case
If X is a discrete random variable with a countable number of countable outcomes {\displaystyle x_{1},x_{2},\ldots ,x_{k}} with probabilities {\displaystyle p_{1},p_{2},\ldots ,p_{k},}. The expected value is defined as follows.
{\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}.}
b] Absolute continuous case
Consider X to be a random variable possessing the probability density function f (x), then the value of expectation is termed as the Lebesgue integral given by
{\displaystyle \operatorname {E} [X]=\int _{\mathbb {R} }xf(x)\,dx,}
c] General case
X is a random variable interpreted on a probability space {\displaystyle (\Omega ,\Sigma ,\operatorname {P} )},, then the expectation of X that is represented by E (X) is given by
{\displaystyle \operatorname {E} [X]=\int _{\Omega }X(\omega )\,d\operatorname {P} (\omega ).}
Expected Value Properties
The basic properties of the expected value of a random variable are as follows.
a] In the case of a general random variable X,
b] Assume 1A denotes the characteristic function of a specific event A, then {\displaystyle \operatorname {E} [{\mathbf {1} }_{A}]=1\cdot \operatorname {P} (A)+0\cdot \operatorname {P} (\Omega \setminus A)=\operatorname {P} (A).}
c] Formulae in terms of cumulative distribution function.
If F (x) is the cumulative distribution function (CDF) of the measure of probability P and X is a random variable, then
{\displaystyle \operatorname {E} [X]=\int _{\overline {\mathbb {R} }}x\,dF(x),}
In addition to that, {\displaystyle \displaystyle \operatorname {E} [X]=\int \limits _{0}^{\infty }(1-F(x))\,dx-\int \limits _{-\infty }^{0}F(x)\,dx,}
d] Non-negativity: \text \ If \ {\displaystyle X\geq 0} \ (a.s.), \ then \ {\displaystyle \operatorname {E} [X]\geq 0}.
e] Expectation linearity
The operator of expectation is linear in nature, for the two random variables X and Y with a constant a, then {\displaystyle {\begin{aligned}\operatorname {E} [X+Y]&=\operatorname {E} [X]+\operatorname {E} [Y],\\\operatorname {E} [aX]&=a\operatorname {E} [X],\end{aligned}}}
f] In general, the expected value is non-multiplicative in nature.
g] The law of unconscious statistician is used to compute the expected value of a function g (X) of a random variable X when the probability distribution of X is known but not g (X).
{\displaystyle \operatorname {E} [g(X)]=\int _{\mathbb {R} }g(x)f(x)\,dx.}
Expected Value Formula
The expected value can be found using the following formula:
E (X) = P (X) * n
Where:
P (X) – the probability associate with the event
n – the number of the reiterations of the event
Expected Value Examples
Example 1: If X is a random variable that follows Bernoulli distribution with a parameter p, then find the expected value of X.
Answer:
The range of Bernoulli distribution is {0, 1}.
E (X) =0 \cdot P_{X}(0)+1 \cdot P_{X}(1) \\ =0 \cdot(1-p)+1 \cdot p \\ =p
Example 2: When a fair dice is rolled, 1 dollar is paid for the odd number outcome and 2 dollars for an even number outcome. What is the expected value of the amount that is obtained for a roll of dice?
Answer:
When a dice is rolled, the sample space is given by {1, 2, 3, 4, 5, 6}.
The table given below demonstrates the distribution of probability for a roll of dice and the respective money paid for each end result.
| X | 1 | 2 | 3 | 4 | 5 | 6 |
| Probability | 1 / 6 | 1 / 6 | 1 / 6 | 1 / 6 | 1 / 6 | 1 / 6 |
| Money | 1 | 2 | 1 | 2 | 1 | 2 |
Using the expected value formula,
E(x) =1\left(\frac{1}{6}\right)+2\left(\frac{1}{6}\right)+1\left(\frac{1}{6}\right)+2\left(\frac{1}{6}\right)+1\left(\frac{1}{6}\right)+2\left(\frac{1}{6}\right) \\ =\frac{1}{6}+\frac{2}{6}+\frac{1}{6}+\frac{2}{6}+\frac{1}{6}+\frac{2}{6} \\ =\frac{9}{6} \\ =1.5
Example 3: If X is a random variable that follows a Poisson distribution with a parameter λ, then find the expected value of X.
Answer:
The probability mass function of Poisson distribution is given by
P_{X}(k)=\frac{e^{-\lambda} \lambda^{k}}{k !}, \quad \text { for } k=0,1,2, \ldots
The expected value is computed as follows.
E (X) =\sum_{x_{k} \in R_{X}} x_{k} P_{X}\left(x_{k}\right) \\ =\sum_{k=0}^{\infty} k \frac{e^{-\lambda} \lambda^{k}}{k !} \\ =e^{-\lambda} \sum_{k=1}^{\infty} \frac{\lambda^{k}}{(k-1) !} \\ =e^{-\lambda} \sum_{j=0}^{\infty} \frac{\lambda^{(j+1)}}{j !} \quad(\text { by letting } j=k-1) \\ =\lambda e^{-\lambda} \sum_{j=0}^{\infty} \frac{\lambda^{j}}{j !} \\ \left.=\lambda e^{-\lambda} e^{\lambda} \quad \text { (Taylor series for } e^{\lambda}\right) \\ =\lambda
Source: https://probabilityformula.org/expected-value/
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