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The specific question that I'm having trouble with is:

Let $F$ be the CDF of a continuous random variable, and let $f = F^{'}$.

Let function $h(x) = \frac{1}{2}f(-x) + \frac{1}{2}f(x)$. Show that $h$ is a valid PDF.

My instructor provided a solution as follows:

$\int_{-\infty}^{\infty}h(x)dx = \frac{1}{2}\int_{-\infty}^{\infty}f(-x)dx\ +\ \frac{1}{2}\int_{\infty}^{\infty}f(x)dx\ =\ 1$

I'm not sure how he derived this. I attempted to solve it via symmetry, but I'm not sure if it is applicable here, as there is no information about $f(x)$.

How did this result come to be?


Thank you.

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Since $f$ is a PDF, it integrates to one (i.e., $\int_{-\infty}^{\infty}f(x)dx=1$) by definition.

Can you prove that $\int_{-\infty}^{\infty} f(-x)dx=1$ also?

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