Mathematical Expectation of Common Distributions

Mathematical Expectation of Common Distributions

Two-Point Distribution

Binomial Distribution

Proof: $E(X) = \sum_{k=0}^n k C_n^k p^k(1-p)^{n-k} = \sum_{k=1}^n k \frac{n!}{k!(n-k)!} p^k(1-p)^{n-k}$ $= np \sum_{k=1}^n \frac{(n-1)!}{(k-1)!(n-k)!} p^{k-1}(1-p)^{n-k} = np \sum_{j=0}^{n-1} C_{n-1}^j p^j(1-p)^{n-1-j} = np$

Poisson Distribution

Proof: $E(X) = \sum_{k=0}^{\infty} k \frac{\lambda^k}{k!} e^{-\lambda} = \lambda e^{-\lambda} \sum_{k=1}^{\infty} \frac{\lambda^{k-1}}{(k-1)!} = \lambda e^{-\lambda} e^{\lambda} = \lambda$

Uniform Distribution

Proof: $E(X) = \int_a^b x \cdot \frac{1}{b-a} dx = \frac{1}{b-a} \cdot \frac{x^2}{2}\big|_a^b = \frac{b^2-a^2}{2(b-a)} = \frac{a+b}{2}$

Exponential Distribution

Proof: $E(X) = \int_0^{\infty} x \cdot \lambda e^{-\lambda x} dx = \lambda \int_0^{\infty} x e^{-\lambda x} dx$ $= \lambda \left[ -\frac{x}{\lambda} e^{-\lambda x} - \frac{1}{\lambda^2} e^{-\lambda x} \right]_0^{\infty} = \frac{1}{\lambda}$

Normal Distribution

Proof: $E(X) = \int_{-\infty}^{+\infty} x \cdot \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{(x-\mu)^2}{2\sigma^2}} dx$ Let $y = x - \mu$; then: $E(X) = \int_{-\infty}^{+\infty} (y + \mu) \cdot \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{y^2}{2\sigma^2}} dy$ $= \int_{-\infty}^{+\infty} y \cdot \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{y^2}{2\sigma^2}} dy + \mu \int_{-\infty}^{+\infty} \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{y^2}{2\sigma^2}} dy$ The first term is 0 (an odd function), and the second is $\mu$, so $E(X) = \mu$.

Exercises

Exercise 1

Given that the random variable $X$ follows an exponential distribution with parameter $\lambda$, find $E(X)$.

Exercise 2

Given that the random variable $X$ follows a binomial distribution with parameters $n, p$, find $E(X)$.