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Let $\phi$ and $\Phi$ be the standard normal density and distribution functions.

Show that $\Phi(\lambda x) = 1 - \Phi(-\lambda x)$ and that $f(x) = 2\phi(x)\Phi(\lambda x)$ is a probability density function.


My thoughts: the first part is true by symmetry around 0. I tried to show this using the definition of $\phi$ and integral $\Phi$ but struggled. The second part, I'm not sure what is meant by show it is a PDF. Isn't the only criterion that $\int f(x) = 1$? I tried using the expression from the first part with the definition/integrals again, but couldn't seem to get the trick involved.

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Hint. $$\int_{-\infty}^\infty f(x)\,dx = \int_0^\infty (f(x)+f(-x))\,dx $$

(The criterion for being a PDF is (a) that the function is nonnegative, and (b) that its integral is 1).

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