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.

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

Any help would be much appreciated.


1 Answer 1


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.

  • $\begingroup$ 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

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