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$f(x)$ and $F(x)$ are the pdf and cdf of a distribution such that f'(x) exists for all $x$. Let the distribution $g(y) = f(y)/F(b)$ for $-\infty < y < b$ have mean $-f(b)/F(b)$ for all real $b$. Prove that $f(x)$ is a pdf of a standard normal distribution.

Attempt: I get $-f(b)= \int_{-\infty}^b y f(y) \,dy$ by definition of expected value. I am stuck here.

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Picking up where you left off... Differentiate both sides to obtain $-f^\prime(b)=b f(b)$ or $(\log f(b))^\prime=-b$. Integrating again we get $\log f(b)=-b^2/2+c$ for some constant $c$, that is, $f(b)=e^{-b^2/2} e^c$. The constant $c$ is chosen to make $f$ a density function, giving $f(b)={1\over \sqrt{2\pi}}\, e^{-b^2/2}.$