I have $X_1, X_2, .., X_n$ i.i.d. standard normal random variables. If $\overline{X}_n$:

$$\overline{X}_n = \frac{X_1 + X_2 + .. + X_n}{n}$$

What is $Var(\overline{X}_n)$ in terms of n?

The solution steps is:

(1) As $X_1 + X_2 + .. + X_n \sim N(0,n)$

(2) We multiply by $1/n$ we get $N(0,1/n)$

(3) as scaling a random variable with a constant $a$ scales its variance by $a^2$. Therefore $\overline{X}_n$ is a gaussian random variable with mean $0$ and variance $1/n$

My problems:

  • Can you just multiply or add to a distribution? For example, is it right to do: $$N(0,1) * k = N(0,k)$$


$$N(0,1) + r = N(r,1+r)$$

Or it depends on the distribution? When can you and when you can't do this kind of computation with a distribution?

  • I understand that $Var(X*a) = a^2*Var(X)$. But I don't understand how this apply to (3).

I tried to do the next integral with wolframalpha, but it tells me "standard computation time exceeded" so I guess it's a fairly complex integral:

$$ Var(\overline{X}_n)$$ $$= E[\overline{X}_n^2] - (E[\overline{X}_n])^2$$ $$= \displaystyle \int_{-\infty}^{\infty} x^2*\frac{1}{\sigma\sqrt{2\pi}} \exp\left( -\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^{\!2}\,\right) dx - \left( \int_{-\infty}^{\infty} x*\frac{1}{\sigma\sqrt{2\pi}} \exp\left( -\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^{\!2}\,\right)dx \right)^2$$


1 Answer 1

  1. If you have $k$ independent normally distributed random variables, then their sum is also a normal random variable with mean equal to the sum of the means and the variance equal to the sum of the variances: wiki. In your notation, $N(0, 1) + N(0, 1) + \ldots + N(0,1) \sim N(0, k)$. I do not recommend simplifying the notation as you did ($N(0, 1) * k$), because it means absolutely different thing. However, if the variables are not indepedent, then the result does not hold anymore. In other words, you can sum in this way only normal random variables that are independent.

  2. As for multiplying, you can multiply any random variable $Y$ by a scalar $a$, and you will get that its mean is multiplied by $a$ and its variance (if it exists) is mutliplied by $a^2$.

  3. As for adding a constant $C$, you can regard this constant $C$ as a degenerate random variable with mean $C$ and zero variance. The result from 1. can be extended in this case, which means that in your notation (however, I would not recommend using it), $N(0, 1) + r \sim N(r, 1)$.

  4. In your sketch of the proof, (3) directly follows from (1) and (2), you do not need to use any properties of the variance for it.


Not the answer you're looking for? Browse other questions tagged or ask your own question.