Can This Integral Be Solved Using Either A Stochastic Approach or As Structured Assuming Lognormality?

Question

I am interested in identifying the expected value of a negative exponential function:

$1-e^{-pw(x-r)}$

where x is a lognormal variable so that the expected value of the function is:

$\int{[1-e^{-pw(x-r)}]\frac{-e^{-\frac{1}{2}\frac{(log(x)-\mu)^2}{\sigma^2}}}{x \sigma\sqrt{2\pi}}dx}$

Since x is effectively a Brownian motion with known terminal T (so the only issue is uncertainty), can I structure the integral differently to get the expected value of the function?

I've tried looking at the problem in the following way, assuming that x follows a Brownian motion:

$\int{[1-e^{-pw((\mu-r)t + \sigma x)}]\frac{-e^{-\frac{1}{2}\frac{x^2}{t}}}{\sqrt{t}\sqrt{2\pi}}dx}$

but am not sure that this is the right way to think about solving the problem.

I used a Taylor series approximation to solve the first integral, but it's a poor approximation for the range of values I am considering for p and t (embedded in $\mu$ and $\sigma$ in the first formulation).

Answer 1

There are all sorts of extraneous variables there, so I'll write

$\begin{array}\\ \int{[1-e^{-pw((\mu-r)t + \sigma x)}]\frac{-e^{-\frac{1}{2}\frac{x^2}{t}}}{\sqrt{t}\sqrt{2\pi}}dx} &=\int{[1-e^{-at + s x)}]\frac{-e^{-\frac{x^2}{2t}}}{\sqrt{2\pi t}}dx}\\ &=\int\frac{-e^{-\frac{x^2}{2t}}}{\sqrt{2\pi t}}dx -\int{e^{-at + s x}\frac{-e^{-\frac{x^2}{2t}}}{\sqrt{2\pi t}}dx}\\ &=-\frac1{\sqrt{2\pi t}}\int e^{-\frac{x^2}{2t}}dx +\frac{e^{-at}}{\sqrt{2\pi t}}\int e^{-\frac{x^2}{2t}+sx}dx\\ \end{array} $

These are both versions of the normal distribution, with the second having its peak shifted away from zero.

You should be able to take it from here.