2
$\begingroup$

I was reading about the Gaussian shift theorem in "An Introduction to Exotic Option Pricing" by Peter Buchen and came across a question that I can't seem to figure. In the book, he uses F(Z) (a measurable scalar function of Z, Z being Gaussian rv with a normal variate) but the function doesn't appear in the question and rather just uses Z1 and Z2.

enter image description here

where 1D is the univariate Gaussian distribution and GST is the Gaussian shift theorem

Any help would be much appreciated.

$\endgroup$

1 Answer 1

1
$\begingroup$

The Gaussian Shift Theorem says that, for a standard Gaussian random variable $Z$, constant $c$, and function $F$, we have the expectation \begin{align*} E\left(e^{cZ} F(Z) \right) = e^{\frac{1}{2} c^2}E\big(F(Z+c) \big). \end{align*} Given the decomposition of $Z_2=\rho Z_1 + \sqrt{1-\rho^2} Z$, where $Z$ is independent of $Z_1$, \begin{align*} E\left(Z_1 e^{a Z_2} \right) &= E\left(Z_1 e^{a \rho Z_1 + a \sqrt{1-\rho^2}Z } \right)\\ &=E\left(Z_1 e^{a \rho Z_1}\right) E\left(e^{a \sqrt{1-\rho^2}Z } \right). \end{align*} Now, you can apply the Gaussian Shift Theorem to compute each of them.

$\endgroup$
4
  • $\begingroup$ My attempt gave me: e^(0.5a^2)E(Z1+aρ)e^(0.5(a^2-a^2ρ^2))E(a√(1−ρ^2) +Z) = a^2e^(0.5a^2) Should E(a√(1−ρ^2) +Z) not have been there to begin with if E(e^(a√(1−ρ^2)Z)) is treated as the expectation of a constant? $\endgroup$ Commented Sep 9, 2018 at 12:51
  • $\begingroup$ My bad, my attempt gave me √(1−ρ^2)a^2pe^(0.5a^2) not a^2e^(0.5a^2), and that I assumed E(Z1) and E(Z) would both equal 1 if they follow the Gaussian distribution. $\endgroup$ Commented Sep 9, 2018 at 13:33
  • $\begingroup$ assuming $c= a \rho$, for the first one, and $a=a\sqrt{1-\rho^2}$, for the second one. $\endgroup$
    – Gordon
    Commented Sep 9, 2018 at 13:55
  • $\begingroup$ I'm afraid I don't completely follow. $\endgroup$ Commented Sep 12, 2018 at 5:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.