(a) If $X_{i}\sim\mathcal{N}(\mu_{i},\sigma^{2}_{i})$ are independent and $a_{i}\in\mathbb{R}$, $1\leq i \leq n$, then \begin{align*} Y = \sum_{i=1}^{n}a_{i}X_{i}\sim\mathcal{N}\left(\sum_{i=1}^{n}a_{i}\mu_{i},\sum_{i=1}^{n}a^{2}_{i}\sigma^{2}_{i}\right) \end{align*}

(b) Prove that $\overline{X} = \displaystyle\frac{1}{n}\sum_{i=1}^{n}X_{i}\sim\mathcal{N}\displaystyle\left(\mu,\frac{\sigma^{2}}{n}\right)$, where $X_{i}\sim\mathcal{N}(\mu,\sigma^{2})$

(c) Suppose that $\displaystyle U = \sum_{i=1}^{n}\left(\frac{X_{i}-\mu}{\sigma}\right)^{2}\sim\chi^{2}_{(n)}$, where $X_{i}\sim\mathcal{N}(\mu,\sigma^{2})$.


(a) We first need to remember that $M_{a_{i}X_{i}}(t) = M_{X_{i}}(a_{i}t)$. Based on this as well as on the independence of the random variables, we have \begin{align*} M_{Y}(t) & = \prod_{i=1}^{n}M_{X_{i}}(a_{i}t) = \exp\left[\sum_{i=1}^{n}\left(\mu_{i}a_{i}t +\frac{a^{2}_{i}t^{2}\sigma^{2}_{i}}{2}\right)\right]\\\\ & = \exp\left(t\sum_{i=1}^{n}a_{i}\mu_{i} + \frac{t^{2}}{2}\sum_{i=1}^{N}a^{2}_{i}\sigma^{2}_{i}\right) \end{align*}

where the last expression is the moment generating function of the corresponding normal distribution presented above.

(b) This is a particular case from the application of (a). It suffices to take $a_{i} = 1/n$.

(c) We know that $\displaystyle Z = (X-\mu)/\sigma\sim\mathcal{N}(0,1)$. Moreover, we do also know that $Z^{2}\sim\chi^{2}_{(1)}$ (which can be proved using moment generating functions). Since the sum of independent chi-squared distributions is also a chi-squared distribution, the result follows.

  • $\begingroup$ Solution seems decent to me. Maybe it's worth mentioning in part (c) that the sum of Independent chi squared variables is again chi squared (you're missing the word independent) $\endgroup$ – Alexandros Feb 6 at 18:44
  • $\begingroup$ Thanks for the comment. I have already edited it. $\endgroup$ – APC89 Feb 6 at 18:45

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