If two random variables have the same variance and expectation, do they have the same distribution? If two random variables say $X$ and $Y$ have the same variance and expectation, do they have the same distrubution?
 A: No.
Example 1: Let $X$ be binomially distributed with $n = 4$ and $p = 1/2$; then $\mu = 2$ and $\sigma^2 = 1$. Let $Y$ be normally distributed with $\mu = 2$ and $\sigma^2 = 1$. Clearly these do not have the same distribution; one is discrete and the other is continuous.
Example 2: Let $X$ be exponentially distributed with $\mu = 10$ ($\implies \sigma = 10$). Let $Y$ be normally distributed with $\mu = 10$ and $\sigma = 10$. These do not have the same distribution, since $Y$ can assume negative values but $X$ can't. 
These examples feel kind of cheap because a normal distribution has its mean and variance (or standard deviation) as its parameters, so for a slightly more concrete example:
Let $$X = \begin{cases} 1,& \textrm{w/ probability 1/2} \\ -1, & \textrm{w/ probability 1/2}  \end{cases}$$
and let $$Y = \begin{cases} 2,& \textrm{w/ probability 1/8} \\ 0, & \textrm{w/ probability 3/4} \\ -2, & \textrm{w/ probability 1/8.}  \end{cases}$$
One can compute for both $X$ and $Y$ that $\mu = 0$ and $\sigma^2 = 1$.
A: The following two distributions both have expected value $0$ and variance $1$:


*

*the $N(0,1)$ distribution

*the uniform distribution on $\left[-\sqrt 3, +\sqrt 3\,\,\right]$

