It's a little tricky to me, I am not sure how to proceed...my atempt was to stick strictly to the definitions given and the formula for the (continuous) expected value.

Prove that $E[e^{\mu+\sigma Z}\mathbb{1}_{Z>-d}]=e^{\mu+\frac{\sigma ^2}{2}}\Phi(d+\sigma)$, where $\Phi$ is the standard normal distribution function, $Z \sim N(0,1)$(i.e. standard normal distribution) and $\mathbb{1}_{Z>-d}$ denotes the following indicator function of an event such that,

$$\mathbb{1}_{Z>-d}=1 \text{ when } Z >-d \text{ and, } \mathbb{1}_{Z>-d}=0 \text{ otherwise}$$

during your work, denote the standard normal density function as $\phi$.

I only got up to thinking $e^{\mu+\sigma Z}\mathbb{1}_{Z>-d}$ is essentially just $e^{\mu+\sigma Z}$ if $Z>-d$ for some $d$ and $0$ otherwise. And so, the expected value is defined (for continuous random variables) $E[X]= \int_{-\infty}^{\infty}xf(x)dx$ where $f(x)$ is the probability for each random variable $x$. I thought this would lead me to using $\phi$ and ultimately leading me to $\Phi$ in my final conclusion as in the RHS.

But I am not sure about how to proceed, would someone please help me? It would be great if you could explain the steps to me...thank you very much in advance...


$\textbf{hint}$ $$ \int^{-d}_{-\infty} 0\cdot \mathrm{e}^{\mu + \sigma Z}\mathrm{e}^{-\frac{Z^2}{2}}dZ + \int_{-d}^{\infty}1\cdot \mathrm{e}^{\mu + \sigma Z}\mathrm{e}^{-\frac{Z^2}{2}}dZ $$ can you integrate this?

$$ \mu + \sigma z-\frac{z^2}{2} = -\frac{z^2-2\sigma z}{2}+\mu=-\frac{(z-\sigma)^2}{2}+\frac{\sigma^2}{2}+\mu $$ thus the integral becomes $$ \mathrm{e}^{\frac{\sigma^2}{2}+\mu}\int_{-d}^{\infty}\mathrm{e}^{-\frac{(z-\sigma)^2}{2}}dz $$

  • $\begingroup$ That looks like it might get me to the RHS! I'll try.... $\endgroup$ – Kydo Oct 29 '15 at 22:19
  • $\begingroup$ I get a polynomial power so do I use integretion by parts here...? $\endgroup$ – Kydo Oct 29 '15 at 22:22
  • $\begingroup$ Thanks for a massive hint. So I suspect I should use an interchange of variables...I set $y=z-\sigma$ which I get $dy=dz$ so by substitution, I get something similar. My one last question is...since $\Phi(x)=\frac{1}{\sqrt{2 \pi}}\int_{-\infty}^xe^{- \frac{y^2}{2}} dy$, how do I get the coefficient $\frac{1}{\sqrt{2 \pi}}$ from the form you've derived? $\Phi(d+ \sigma)=\frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{d+ \sigma}e^{- \frac{y^2}{2}} dy$ is what I need, right? Sorry to ask so much! $\endgroup$ – Kydo Oct 30 '15 at 0:21
  • $\begingroup$ Sorry about the delay. The coefficient comes from the normal distribution that I am integrating with (I didn't add it because i was focused on the integral). $\endgroup$ – Chinny84 Oct 30 '15 at 1:00
  • $\begingroup$ Ah, okay, that makes it clear, thanks! $\endgroup$ – Kydo Oct 30 '15 at 11:43

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