I consider $n$ independent random variables with $$X_1 \sim N(\mu_1, \sigma_1^2),\ldots,X_n \sim N(\mu_n, \sigma_n^2).$$ I was able to show that $$ \sum_{i=1}^n a_i X_i +b \sim N(\sum_{i=1}^n a_i \mu_i+b, \sum_{i=1}^n a^2_i \sigma_n^2 ).$$

I managed to do that using moment-generating functions $X_1+X_2 \sim N(\mu_1+\mu_2, \sigma_1^2 +\sigma_2^2) $ and $M_{X_1+X_2}(t) = exp(\mu_1+\mu_2)t + \frac{(\sigma_1^2 +\sigma_2^2) \cdot t}{2}) $

How can I show now that $\bar{X}\sim N( \mu, \frac{\sigma^2}{n})$ Can somebody give me a hint?

  • 1
    $\begingroup$ $\bar X = \frac1n \sum X_i$ so you can use the moment generating function again $\endgroup$
    – Henry
    Aug 5, 2019 at 10:38
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    $\begingroup$ Ok I thought I would be easier:( $\endgroup$
    – Leon1998
    Aug 5, 2019 at 10:46

1 Answer 1


You can just use the previous result twice:

First derive the distribution of $Y= \sum_i X_i$ and then the one of $Z = \frac{1}{n}Y$, both are particular cases of the first result.


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