the question is:

$X, Y$ are i.i.d and standard normal. $\theta$ $\epsilon $ $(0, 1)$. Define a P-equivalent measure Q by the Radon Nikodym derivative $\frac{dQ}{dP} := \frac{e^{XY\theta}}{Ee^{XY\theta}}$. What is the joint distribution of X and Y under Q? Are they also independent under Q?

How do I go about this? How do I use the change of measure? Do moment generating functions come into play? I tried defining $Z := X + Y$ and computing $E^Qe^{Z}$ but I am unable to glean any insights from this.

Thank you.


Let $$ p(x, y) \equiv \frac{1}{\sqrt{2\pi}}exp\left(-\frac{x^2}{2}\right)\times\frac{1}{\sqrt{2\pi}}exp\left(-\frac{y^2}{2}\right)$$be the joint standard normal distribution. Let $q(x, y)$ be the density function under Q. Let $L$ be the Lebesgue measure.

Then by definition $$\frac{dP}{dL}=p(x, y)\quad\text{and}\quad \frac{dQ}{dL}=q(x, y).$$

Hence $$ q(x, y) = \frac{dQ}{dL} = \frac{dQ}{dP}\frac{dP}{dL} = \frac{e^{xy\theta}}{Ee^{XY\theta}} p(x, y).$$

One can verify $e^{xy\theta}$ is not product-term separable function for $\theta \in (0, 1)$. Further note $p(x, y)$ is product-term separable and $Ee^{XY\theta}$ is a function of $\theta$ (considered a constant). Hence $q(x, y)$ is not product-term separable. Hence $X, Y$ are not independent under probability measure $Q$.

PS: Noting the moment generation function for a standard normal distribution is $M(t)=e^{\frac{1}{2}t^2}$, $$Ee^{XY\theta}=E_XE_Ye^{(X\theta)Y}=E_Xe^{\frac{1}{2}\theta^2X^2} = \int \frac{1}{\sqrt{2\pi}}exp\left(-\frac{(1-\theta^2)x^2}{2}\right)=\frac{1}{\sqrt{1-\theta^2}}.$$

For separability, one can refer to:theorem 3. If you want to avoid separability, you can calculate the marginal densities and show the joint density is not a product of marginals.

| cite | improve this answer | |
  • $\begingroup$ thank you so much for the insight. Can you also elaborate on how you calculate that expectation? I'm not sure how you break it up into $E_X$ and $E_Y$. $\endgroup$ – arnavlohe15 Nov 1 '19 at 21:16
  • $\begingroup$ It is based on Fubini's theorem. The first integration of $E$ is over the joint distribution, and then it is split into two integrations first over $Y$, then over $X$. When computing $E_Y$, I used the moment generation function formula. $\endgroup$ – Xiaohai Zhang Nov 1 '19 at 21:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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