Given a sequence of independent random variable $X_1,X_2,\dots,X_n$ with $f_X$, how to find $f_A$ if $A = S_n/n$ is their average

Question

My book gives the following. For a standard normal distribution: $$f_X(x)=\frac{1}{\sqrt{2\pi}}e^{-x^2/2}$$ The density of the sum of n independent standard normal random variable ($S_n$) is: $$f_{S_n}(x)=\frac{1}{\sqrt{2\pi n}}e^{-x^2/2n}$$ Similarly for the exponential distribution $$f_X(x)=\lambda e^{-\lambda x}$$ The density of the sum of n independent random variable each with the same exponential density is: $$ f_{S_n}(x)=\frac{\lambda e^{-\lambda x}(\lambda x)^{n-1}}{(n-1)!}$$ Where $S_n = X_1 + X_2 +...X_n$. If $A = S_n/n$ and I want to find $f_A$ for the standard normal and exponential density $f_X(x)= e^{-x}$, is the following approach correct: $$F_A(a) = P(S_n/n < a) = F_{S_n}(na)$$ $$dF_A(a)/da = f_A(a) = nf_{S_n}(na)$$

From this, $f_A$ for the standard normal case: $$f_{A}(a)=\frac{\sqrt{n}}{\sqrt{2\pi}}e^{-na^2/2}$$

$f_A$ for the exponential case: $$ f_{A}(a)=\frac{n^ne^{-na}x^{n-1}}{(n-1)!}$$

This doesn't match the solution in my book below:

Answer 1

The book is wrong.

Normal distribution. For large $n$ the distribution should approach a $\delta$ function, which yours does. The book spreads out to be flat.

Exponential distribution. You have a minor error, using $x$ when you mean $a$. For $n=1$, your answer is correct, while the book has an extra factor of $x$.