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Suppose $f(x)$ and $F(x)$ is p.d.f and c.d.f of continuous random variable $x$ such that $f'(x)$ exist for every real value $x$. Suppose that the mean of $Y$ with p.d.f

g(y)=$\begin{cases} f(y)/F(b) , -\infty < y < b \\ 0, b<y<\infty \end{cases}$

equal to -f(b)F(b) for every real number $b$. Show that $f(x)$ is a p.d.f of standard normal distribution

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  • $\begingroup$ Do you know a simple differential equation satisfied by $\phi$ (the standard normal PDF)? $\endgroup$
    – Ian
    Commented Dec 9, 2017 at 15:58
  • $\begingroup$ i am sorry, i don't $\endgroup$
    – cavvot
    Commented Dec 9, 2017 at 15:59
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    $\begingroup$ I get stuck with the differential equation $b f(b) = -f'(b)F(B)^2 -2f(b)^2 F(b)$ $\endgroup$
    – cavvot
    Commented Dec 9, 2017 at 16:00

1 Answer 1

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It looks like there is a mistake somewhere. If this were true, we would have $$xf(x) = -f'(x)F(x)^2 - 2f(x)^2F(x) \quad, \forall x \in \mathbb{R}$$
Let's see if $f(x) = c e^{-x^2/2}$ satisfies this (with $c = \frac{1}{\sqrt{2\pi}}$). Substituting $x=0$ gives: $$ 0 = -f'(0)F(0)^2 - 2f(0)^2F(0) $$ But $F(0)=1/2$, $f'(0)=0$, $f(0)=c$, so we get $$ 0 = -c^2$$ So $c=0$, a contradiction.

I think the given information is stated incorrectly. It should be stated as $$ \int_{-\infty}^b x f(x)dx = -f(b) \quad, \forall b \in \mathbb{R} $$ This yields an ODE that is solved by a Gaussian. This modified information is like saying that the mean of $Y_b$ is equal to $-f(b)/F(b)$ (rather than $-f(b)F(b)$) for all $b$.

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