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Consider the integral of a normal distribution: $$\int_a^b f(x)\,\mathrm d x=c $$

and a second integral for the expected value: $$ \int_a^b x\cdot f(x)\,\mathrm dx $$ Since you know the first integral is equal to $c$, what is a good way to evaluate the second integral to find the expected value? Integration by parts doesn't seem to work.

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up vote 4 down vote accepted

$$ \int_a^b \frac1c f(x)\,dx = 1. $$ Therefore $\dfrac1c f(x)$ is a probability density function on the interval $[a,b]$. It is the conditional probability density of a normal random variable given that it's in that interval. So $$ \int_a^b \frac1c x f(x)\,dx $$ is the conditional expected value of a normal random variable given that it's in that interval.

Later note: To see why this is the conditional distribution, consider the problem of finding the conditional probability $\Pr(X\in S\mid a\le X\le b)$, where $X$ is a random variable whose density is $f$: $$ \Pr(X\in S\mid a\le X\le b) = \frac{ \Pr(X\in S\text{ and }a\le X\le b ) }{\Pr(a\le X\le b)} $$ $$ = \frac{ \Pr(X\in S\text{ and }a\le X\le b ) }{c} = \frac{\int_{S\cap [a,b]} f(x)\,dx}{c} = \int_{(\text{the relevant set})} \frac1c f(x)\,dx. $$

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