PDF of the exponential of the sum of $N$ independent gaussian random variables

Question

Is it true that the PDF of the sum of independent gaussian and identical random variables is gaussian with mean $N\cdot\mu$ and variance $N \cdot \sigma^2$? And if so how do we get the PDF of the exponential of such a sum?

Answer 1

Let $x$and $y$ be Gaussian random variables; $x \sim \mathcal{N}(x; \mu_x, \sigma_x^2)$ and $y \sim \mathcal{N}(y; \mu_y, \sigma_y^2)$.

We know the Gaussian random variable's PDF is an exponential function. For instance:

$$ p(x) = \frac{e^{-\frac{1}{2\sigma_x^2}(x-\mu_x)^2}}{\sqrt{2\pi} \sigma_x}, $$

and similarly for $p(y)$.

Then, consider the sum $z = x + y$.

PDF of $z$ is a joint PDF of $x$ and $y$ and by independence,

$p(z) = p(x,y) = p(x)\cdot p(y)$.

We see then $p(z)$ will be a multiplication of two exponentials $p(x)$ and $p(y)$ and you know that exponents will add up to form a new exponential function.

By performing the completion of squares, you will get a new PDF that looks just like a Gaussian PDF with mean $\mu_z$ and the variance $\sigma_z^2$.

Now, for independent gaussian, these can be computed as follows:

$$ \mu_z = \mathbb{E}[z] = \mathbb{E}[x + y] = \mathbb{E}[x] + \mathbb{E}[y] = \mu_x + \mu_y, $$ $$ \sigma_z^2 = var[z] = var[x + y] = var[x] + var[y] = \sigma_x^2 + \sigma_y^2. $$

Assume now that $x$ and $y$ are actually identical random variables. Then, the above will be $N\mu$ and $N\sigma^2$ for $N$ identical independent Gaussian random variables.

Then, you can write now the PDF of $z$ just like how you would write the PDF for any Gaussian random variable:

$$ p(z) = \frac{e^{-\frac{1}{2\sigma_z^2}(z-\mu_z)^2}}{\sqrt{2\pi} \sigma_z}, $$

or more explicitly to your question, let $X$ be the sum of independent identical Gaussian random variables $x_i$ for $i = 1$ to $i = N$ so that $X = \sum_{i=1}^{N}{x_i}$ and each $x_i$ having the mean and the variance as $\mu_x$ and $\sigma_x$. $$ p(X) = \frac{e^{-\frac{1}{2N\sigma_x^2}(X-N\mu_x)^2}}{\sqrt{2\pi N} \sigma_x}, $$

Answer 2

Let $S=\sum X_k$. The cdf for $e^S$ can be expressed as $F(x)=P(e^S\le x)=P(S\le ln(x))=\frac{1}{\sqrt{2\pi N}\sigma}\int\limits_{-\infty}^{ln(x)}e^{-\frac{(u-N\mu)^2}{2N\sigma^2}}du$
Thus the pdf $f(x)=\frac{1}{\sqrt{2\pi N}x\sigma}e^{-\frac{(ln(x)-N\mu)^2}{2N\sigma^2}}$