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I need to calculate the following integral

$$\int_{\mu+c}^{\infty} y\cdot \frac{1}{\sigma\sqrt{2\pi}}e^{(y-\mu-w)^2/2\sigma^2}dy$$

So essentially $y\sim N (\mu+w, \sigma^2)$ and im trying to calculate its expected value from $\mu+c$ to $\infty$

I know I will have to use integration by parts, but don't know how to proceed.

More importantly however, when I calculate the integral, will I get an expression which includes $w$? I really need the $w$ in there.

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  • $\begingroup$ The result will definitely be a function of $w$. $\endgroup$ – Glen_b Sep 13 '14 at 2:08
  • $\begingroup$ The integral does not converge: presumably you intend for the argument of $\exp$ to be negated. $\endgroup$ – whuber Sep 13 '14 at 21:45
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Hint: setting $\sigma^2=1$ for convenience, we have that

$$\begin{align} \frac{\mathrm d}{\mathrm dy} \exp(-(y-\mu-w)^2/2) &= -(y-\mu-w)\exp(-(y-\mu-w)^2/2)\\ &= -y\exp(-(y-\mu-w)^2/2 \\ &\qquad + (\mu+w)\exp(-(y-\mu-w)^2/2 \end{align}$$ and so you should be able to write the integrand as $(\mu+w)$ times a Gaussian density less a perfect differential, and thus get the integral to work out to something involving $\Phi(\cdot)$ and an $\exp(g(w))$ term where $g(w)$ is a quadratic function of $w$.

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  • $\begingroup$ i dilip thanks for that. Are you hinting at using integration by parts? If i integrate the expression you'vve provided i will end up getting $exp^(-(y-\mu-w)^2/2)$ rather than the integral of it. Won't i? $\endgroup$ – user55405 Sep 14 '14 at 3:35
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    $\begingroup$ @user55405 I am not hinting at using integration by parts. I am saying outright that $$\begin{align}\int y\exp(-(y-\mu-w)^2/2\,\mathrm dy&=-\int\frac{\mathrm d}{\mathrm dy} \exp(-(y-\mu-w)^2/2)\,\mathrm dy+\int(\mu+w)\exp(-(y-\mu-w)^2/2\,\mathrm dy\\&=-\exp(-(y-\mu-w)^2/2)+(\mu+w)\int\exp(-(y-\mu-w)^2/2\,\mathrm dy\end{align}$$ from which, when you change to definite integrals instead of antiderivatives as in the above expression, you get an exponential in $w$ from the first term and something involving the Gaussian CDF $\Phi(\cdot)$ from the second term on the right. $\endgroup$ – Dilip Sarwate Sep 14 '14 at 11:54
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The conditional expectation for a normal random variable with mean $\tau$ and standard deviation $\sigma$ can be found from this formula, which is available, for example, here: www.actuaries.org/LIBRARY/ASTIN/vol35no1/189.pdf ‎

$$ E[X | X > q ]={\tau + {{\sigma} \phi \left({{q-\tau} \over \sigma} \right) \over {1-\Phi \left( {{q-\tau} \over \sigma}\right)} } },$$

where $\Phi$ is the CDF and $\phi$ is the PDF for the standard normal. The general formula that can be used to derive these type of expectations is given by

$$ E[X | X > q ]={ {\int_q^\infty xf(x)dx} \over {P[X>q]}}$$

Putting these together with your parameters, we have

$$E[Y | Y > \mu+c ]={ {{{\int_{\mu+c}^\infty {y \over {\sigma \sqrt{2\pi}}} e^{{-{\left ( {y-\mu-w} \right)^2}} \over {2 \sigma^2}} dy} }\over {1-\Phi \left( {{c-w} \over \sigma}\right)} }=\mu+w +\left[{\sigma {\phi \left( {c-w} \over \sigma \right) } \over {1-\Phi\left( {c-w} \over \sigma \right)}} \right] }$$

Finally, then, we can solve for the integral you want:

$${\int_{\mu+c}^\infty {y \over {\sigma \sqrt{2\pi}}} e^{{-{\left ( {y-\mu-w} \right)^2}} \over {2 \sigma^2}} dy}=\left(\mu+w \right) \left[{1-\Phi\left( {c-w} \over \sigma \right)} \right]+\sigma {\phi \left( {c-w} \over \sigma \right) }$$

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