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I'm trying to train myself to recognize probability densities by deriving PDFs from bare kernel functions. In other words, I find a constant expression by integrating a kernel function over its support range, and divide the original kernel function by that expression, so that it equals 1 and thus is a PDF.

For the Normal distribution the kernel is: $e^{-{{(x-a)^2} \over b}}$. So, I let $u = {{x-a} \over \sqrt b}$ then $du = {1 \over \sqrt b}$, so...

$$\int e^{-{{(x-a)^2} \over b}}dx = \sqrt b \int e^{-{{(x-a)^2} \over b}}{1 \over \sqrt b}dx = \sqrt b \int e^{-u^2}du = \sqrt b \sqrt \pi$$

Let $b = \sigma^2 , a = \mu$ and we have... $$\sqrt{\sigma^2 \pi} = \int e^{-{{(x-\mu)^2} \over \sigma^2}}dx \Rightarrow 1 = \int {{e^{-{{(x-\mu)^2} \over \sigma^2}}} \over {\sqrt{\sigma^2 \pi}}} dx$$

...but the problem is that ${{e^{-{{(x-\mu)^2} \over \sigma^2}}} \over {\sqrt{\sigma^2 \pi}}}$ is not the Normal PDF! This is the Normal PDF:

$${{e^{-{{(x-\mu)^2} \over 2\sigma^2}}} \over {\sqrt{2\sigma^2 \pi}}}$$

It seems to me that I could have defined $b$ in the original expression to have any coefficient, and would have still satisfied the conditions for a PDF. So, how would I have arrived at the $2\sigma^2$ parameterization if I didn't expect it ahead of time?

Thanks.

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The problem is that you cannot do $b=\sigma^2$, because the kernal is actually $$ e^{\frac{-(x-\mu)^2}{2\sigma^2}} $$

So you should set $b=2\sigma^2$, and you will be fine.

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