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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.

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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.

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  • $\begingroup$ 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

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