0
$\begingroup$

Suppose that $X_1, X_2, X_3$ are i.i.d. normal random variables with mean $0$ and variance $1$. What is the distribution of $\overline{X} = \frac{1}{3}(X_1+X_2+X_3)$?

I don't quite understand how to generalize CLT to multivariate random variables. Any help? Thanks!

$\endgroup$
1
$\begingroup$

This question has nothing to do with the Central Limit Theorem. Nor does it really have anything to do with multivariate normals. All this problem involves is the fact that if $X\sim N(\mu_x,\sigma^2_x)$ and $Y\sim N(\mu_y,\sigma^2_y)$ and $X$ and $Y$ are independent, then it is a nice fact that $$X+Y\sim N(\mu_x+\mu_y,\sigma^2_x+\sigma^2_y)$$ In other words, the sum of two independent normals is again normal, and the means and variances just add. (The first part is still true even if $X$ and $Y$ are dependent, but in that case the variances are no longer additive.)

So you're given $X_i\sim N(0,1)$, told the $X_i$ are independent, and asked for the distribution of $\bar{X}=\frac{1}{3}\sum X_i$. You can use the above fact to find $$\sum X_i\sim N(0,3)$$ Now you just want to rescale this variable by $\frac{1}{3}$. This preserves normality, but multiplies the mean by $\frac{1}{3}$ and the variance by $\frac{1}{9}$. So the final answer is $$\bar{X}\sim N\left(0,\frac{1}{3}\right)$$ Note that nowhere do you have to invoke the Central Limit Theorem, which is required only for approximating the distribution of $\bar{X}$ when the $X_i$ are not normally distributed.

$\endgroup$
1
$\begingroup$

Let $X=[X_1,\ldots,X_n]^T$ be drawn iid from $\mathcal{N}(\mu,\Sigma)$, then $Y=C^TX$ will be drawn from $\mathcal{N}(C^T\mu,C^T\Sigma C)$

In your example, $\Sigma=I_{3\times3}, n=3, C=[\frac{1}{3},\frac{1}{3},\frac{1}{3}]^T$, and $\mu=[0,0,0]^T$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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