$E[X|X>c]$ = $\frac{\phi(c)}{1-\Phi(c)}$ , given X is $N(0,1)$ , how to derive this?

Question

I came across this useful formula that , if $X$ is distributed as a standard normal , then $E[X\mid X>c]$ = $\frac{\phi(c)}{1-\Phi(c)}$. Where $\phi(x)$ is the probability density function of the standards normal and $\Phi(c)$ is the cdf of the standard normal. I was wondering how to derive this formula.

I thought $$E[X\mid X>c]= \frac { \int_{c}^{\infty} x f_X(x) dx }{1-\Phi(c)}$$ so how to get the formula $\frac{\phi(c)}{1-\Phi(c)}$ ?

Answer 1

You're nearly there, now just observe that since $$ \frac{d}{dx}\phi(x)=-x\phi(x)$$ it follows that $$ \int_{c}^{\infty}x\phi(x)\;dx=-\phi(x)\Big|_{c}^{\infty}=\phi(c)$$