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Lads, I need some insight here.

Given our $f(x) = \sqrt{\frac2\pi}e^{-(x^2/2)}$ where x is non-negative, I have to prove that $f(x)$ is a valid pdf.

My work done so far: I've tried to pull out the constant, $\sqrt{\frac2\pi}$, out of the integral, to make the integral look like the normal distribution and I keep getting 2 instead of 1, which is not what I want because I have to get a 1 to prove that it's a valid pdf. What am I missing here?

Apologies for my poor symbols.

George Law
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KevinG
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  • Are you forgetting to use that $x$ is non-negative? This would account for the factor of 2, since $$\frac{1}{2} \int_{-\infty}^\infty \sqrt{\frac{2}{\pi}}e^{-x^2/2},\text{d}x = \int_0^\infty \sqrt{\frac{2}{\pi}}e^{-x^2/2},\text{d}x$$ – Nicholas Stull Nov 17 '16 at 15:43
  • I don't know if that changes anything when you are proving if it's a valid pdf – KevinG Nov 17 '16 at 16:05
  • The reason I said this is I wanted to know what integral you were computing to get 2. If $x$ is non-negative, then you are integrating over $(0,\infty)$, but most standard computations of the Gaussian integral are over $(-\infty,\infty)$. Can you show the process you used to get $2$ so I can more pointedly address your concern? – Nicholas Stull Nov 28 '16 at 15:03

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