1
$\begingroup$

In general, we know that different random variables $X$, $Y$ can have the same expected value, but different variances. A common example for this would be the game of roulette. Two different game strategies can yield the same expected value while the variances of the random variables still differ.

Now I thought about the following situation:

Assume that $E(X) = E(Y)$ (as in the example above), but also $Var(X) = Var(Y)$.

Do there still exist two random variables with different distributions?

  • 0
    A normal distribution $N(0,3)$ and a student-t distribution $t_3$, both have mean $0$ and variance $3$, if that's what you want? – 2017-01-02

1 Answers 1

1

The answer is yes:

Let $X$ be a uniform distribution on $(-1,1)$ and $Y \sim\mathcal N(0,\frac 13)$.

Then $E[X]=E[Y]=0$ and $Var(X)=Var(Y)= \frac 13$. But $X$ and $Y$ do not have the same distribution.

  • 0
    Following my lecture notes, we would actually have to write $Y \sim N(0, \sqrt{1 \over 3})$, otherwise, the variance would be $1 \over 9$. The variance of the normal distribution is always $\sigma^2$. – 2017-01-02
  • 0
    this is just a convention: some write $N(\mu,\sigma)$ others write $N(\mu,\sigma^2)$. – 2017-01-02
  • 0
    But wouldn't this yield a different result? – 2017-01-02
  • 0
    yeah of course, what I meant is that it is a convention whether a normal distributed random variable with mean zero and variance $\frac 13$ is denoted with $N(0,\frac 13)$ or with $N(0, \sqrt{\frac 13})$. Some prefer the first notation, others the second notation. – 2017-01-02