# standard normal expectations

• Nov 24th 2009, 01:58 PM
cribby
standard normal expectations
Calculate $\displaystyle E[e^Z]$, where $\displaystyle Z$ is the standard normal random variable (with density function $\displaystyle f(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}$).

So am I to calculate $\displaystyle \int _{-\infty}^{\infty}e^{f(x)} \cdot f(x) \,\, dx$?

If I'm approaching this correctly, that leaves me with a mess stickier than a barrel of molasses which I don't know how to handle at all. I mean, I know that the integral of f(x) is F(x) where F(x) is the cumulative distribution function for the standard normal distribution, but I don't know what that is explicitly nor how to calculate it, let alone what is being asked here. In lecture, we have used tables of values for $\displaystyle \Phi (x)$, where $\displaystyle \Phi (x)$ (I assume) is this F(x), to approximate definite integrals of f(x) but I do not see how that might help here...so lost.

• Nov 24th 2009, 03:13 PM
qmech
A different interpretation

$\displaystyle \int _{-\infty}^{\infty}e^{x} \cdot f(x) \,\, dx$
• Nov 24th 2009, 03:18 PM
matheagle
Quote:

Originally Posted by qmech

$\displaystyle \int _{-\infty}^{\infty}e^{x} \cdot f(x) \,\, dx$

and instead of completing the square, blah blah, to make this a valid density

NOTE that this is just the MGF evaluated at t=1.
• Nov 25th 2009, 01:19 AM
cribby
Quote:

Originally Posted by matheagle
and instead of completing the square, blah blah, to make this a valid density

NOTE that this is just the MGF evaluated at t=1.

I assure you both, however, that I am stating the problem correctly. In fact, there is another problem I have that requires computing a similar integral (which I can only take as evidence that $\displaystyle E[e^Z]$ is not a typo). To further complicate things, MGFs have not been an addressed topic so I must further assume that a solution is intended to be MGF-free.