# Construct probability measures to get the desired probability distributions of random variables.

Here is a problem from an exam I just took several days ago.

Let $$(\Omega, \mathcal{F}, P)$$ be a probability space.

• Let $$X:\Omega\to\mathbb R$$ be a random variable with $$X>0$$ a.s. and $$EX=1$$. Define $$Q(A)=E[X1_A],\ \ \forall A\in\mathcal{F}.$$ Show that $$Q$$ is a probability measure on $$(\Omega, \mathcal{F})$$ and $$Q\sim P$$, i.e. $$Q< and $$P<.

• Suppose that $$X\sim N(\mu,\sigma^2)$$ under $$P$$ where $$\mu\neq0$$, $$\sigma>0$$ and $$\sigma\neq1$$. Try to construct a probability measure on $$(\Omega, \mathcal{F})$$ such that $$X\sim N(0,1)$$ under $$Q$$.

• Suppose that $$X\sim Poisson(\lambda)$$ under $$P$$ where $$\lambda>0$$ and $$\lambda\neq1$$. Try to construct a probability measure on $$(\Omega, \mathcal{F})$$ such that $$X\sim Poisson(1)$$ under $$Q$$.

The first part is standard and easy for me. But the next two parts stuck me. I have never met and thought those questions before. For the second part, if we introduce $$Y=\frac{X-\mu}{\sigma}$$, then $$Y\sim N(0,1)$$ under $$P$$, which is well-known. But how can I use this and the first part to construct such a probability measure $$Q$$? I cannot move on the third part, too.

Any help would be appreciated.

• Commented Jan 10, 2020 at 16:45
• @MaximilianJanisch Thanks for these links!
– Feng
Commented Jan 10, 2020 at 23:40

For normal case. Let $$f_P(x)=\dfrac{1}{\sigma \sqrt{2\pi}}e^{-(x-\mu)^2/(2\sigma^2)}, \quad f_Q(x)=\dfrac{1}{\sqrt{2\pi}}e^{-x^2/2}$$ be the given and the desired pdfs of $$X$$ under $$P$$ and $$Q$$ respectively. Then define $$\tag{1}\label{1} Q(A) = \mathbb E_P\left[\frac{f_Q(X)}{f_P(X)}I(A)\right].$$ Here index $$P$$ means that we will calculate expectation with respect to initial distribution $$P$$ of $$X$$. Firstly check whether positive r.v. $$Z=\frac{f_Q(X)}{f_P(X)}$$ satisfies conditions from the first descriptive part of the question. We need only to check whether $$\mathbb E_P[Z]=1$$. $$\mathbb E_P\left[\frac{f_Q(X)}{f_P(X)}\right]=\int\limits_{\mathbb R} \frac{f_Q(x)}{f_P(x)} \cdot f_P(x)\, dx=\int\limits_{\mathbb R} f_Q(x)\, dx=1.$$ Check if $$X$$ has standard normal distribution under $$Q$$. For any Borel set $$B$$, one get from \eqref{1} $$Q(X\in B) = \mathbb E_P\left[\frac{f_Q(X)}{f_P(X)}I(X\in B)\right]=\int\limits_B \frac{f_Q(x)}{f_P(x)} \cdot f_P(x)\, dx = \int\limits_{B} f_Q(x)\, dx,$$ so $$f_Q(x)$$ is indeed (standard normal) pdf of $$X$$ under probability measure $$Q$$.
The same you can do with Poisson r.v. $$X$$: $$Q(A) = \mathbb E_P\left[I(A) \cdot \frac{\frac{1}{\not{X!}}e^{-1}}{\frac{\lambda^X}{\not{X!}}e^{-\lambda}}\right],$$ after that $$Q(X=k) = \mathbb E_P\left[I(X=k) \cdot \frac{e^{\lambda-1}}{\lambda^X}\right] = \frac{e^{\lambda-1}}{\lambda^k} P(X=k) = \frac{e^{\lambda-1}}{\lambda^k}\frac{\lambda^k}{k!}e^{-\lambda} = \frac{1}{k!}e^{-1}.$$