Characteristic function of product of two random variables with arbitrary normal distributions

Question

I have $X\sim N(0,5)$ and $Y\sim N(1,1)$ components of a gaussian random vector.

The covariance of $X$ and $Y$ is 2.

I've already proved that $\frac{X}{2}-Y$ is independent from $Y$.

I have to calculate the characteristic function of $\left(\frac{X}{2}-Y\right)\frac{Y}{2}$.

My attempt is the following.

$$\begin{align}\varphi_{\left(\frac{X}{2}-Y\right)\frac{Y}{2}}(\theta)&=\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} e^{i\theta\left(\frac{x}{2}-y\right)\frac{y}{2}}f_{X,Y}(x,y) dxdy\\&=\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}e^{i\theta \frac{xy}{4}}e^{-i\theta\frac{y^2}{2}}f_X(x)f_Y(y)dxdy\\&=\int_{-\infty}^{+\infty}e^{-i\theta\frac{y^2}{2}}f_Y(y)\left(\int_{-\infty}^{+\infty}e^{i\theta \frac{xy}{4}}f_X(x)dx \right)dy\\&=\int_{-\infty}^{+\infty}e^{-i\theta\frac{y^2}{2}}f_Y(y)\varphi_{X}\left(\frac{\theta y}{4}\right)dy\\&=\int_{-\infty}^{+\infty}e^{-i\theta\frac{y^2}{2}}e^{-\frac{1}{2}\left(\frac{5y^2}{16}\theta^2\right)}f_Y(y)dy\end{align}$$

Now, I'm stuck. How can I rewrite that last integral? Maybe I can view it as a Gaussian characteristic function...

Answer 1

I presented the CF of $VW$ with $V\sim N(\mu_1,\sigma_1^2)$ and $W \sim N(\mu_2,\sigma_2^2)$ under different assumptions below.

$V, W$ have the same variance and are independent

Let $V\sim N(\mu_1,\sigma_1^2), W \sim N(\mu_2,\sigma_1^2)$ be two independent normally distributed random variables with the same variances. Then, their product can be written as

$$VW= C \big( Z_1^2-Z_2^2 \big)$$

in which $Z_1$ and $Z_2$ are independent and given by

$$Z_1=\frac{V+W}{\sqrt{2}\sigma_1} \sim N \left (A:=\frac{\mu_1+\mu_2}{\sqrt{2}\sigma_1},1 \right)$$ $$Z_2=\frac{V-W}{\sqrt{2}\sigma_1}\sim N \left (B:=\frac{\mu_1-\mu_2}{\sqrt{2}\sigma_1},1 \right )$$

with $C:=\frac{\sigma_1^2}{2}.$

Note that each of $Z^2_1$ and $Z^2_2$ has a noncentral chi square.

Hence, the characteristic function of $ VW $ is the product of the ones of $ C Z_1^2 $ and $ -C Z_1^2 $, which is

$$\varphi_{VW}(t)=\frac{\exp\left(\frac{iA^2 C t}{1-2iC t}\right)}{(1-2iC t)^{1/2}} \times \frac{\exp\left(\frac{-iB^2 C t}{1+2iC t}\right)}{(1+2iC t)^{1/2}}=\frac{\exp\left(\frac{iA^2 C t}{1-2iC t}-\frac{iB^2 C t}{1+2iC t}\right)}{(1+4C^2 t^2)^{1/2}}. $$

In particular, when $V, W $ follow the standard normal distribution, i.e., $\mu_1=0, \mu_1=0, \sigma^2_1=1, \sigma^2_2=1$, then $A=B=0$, $C=\frac{1}{2}$, and we get the nice formula:

$$\frac 1{\sqrt{1+2t^2}}. $$

$ \fbox{For your specific question:}$, $V=\frac{X}{2}-Y$ and $W=\frac{Y}{2}$, so $\mu_1=-1, \mu_1=\frac{1}{2}, \sigma^2_1=\sigma^2_2=\frac{1}{4}$. Thus, $A=-\frac{1}{\sqrt{2}}$, $B=-\frac{3}{\sqrt{2}}$, and $C=\frac{1}{8}$.

$V, W$ are independent

If $V\sim N(\mu_1,\sigma_1^2), W \sim N(\mu_2,\sigma_2^2)$ are two independent normally distributed random variables, then we can apply the above formula for $VW'$ appearing below:

$$VW=\frac{\sigma_2}{\sigma_1} \times VW'$$

in which $V\sim N(\mu_1,\sigma_1^2), W'= \frac{\sigma_1}{\sigma_2} W\sim N(\frac{\sigma_1}{\sigma_2}\mu_2,\sigma_1^2).$

$V, W$ have the same variances and follow a joint normal distribution (can be dependent)

If $V\sim N(\mu_1,\sigma_1^2), W \sim N(\mu_2,\sigma_1^2)$ are two normally distributed random variables with covariance $\sigma_{12}$, then we can use a similar method to obtain the CF by

$$VW=C_1Z_1^2-C_2Z_2^2$$

with $$A=\frac{\mu_1+\mu_2}{\sqrt{4C_1}}, B=\frac{\mu_1-\mu_2}{\sqrt{4C_2}}, C_1= \frac{\sigma_1^2+\sigma_{12}}{2}, C_2= \frac{\sigma_1^2-\sigma_{12}}{2}.$$

General case: $V, W$ have a joint normal distribution

If $V\sim N(\mu_1,\sigma_1^2), W \sim N(\mu_2,\sigma_2^2)$ are two normally distributed random variables with covariance $\sigma_{12}$, then we can apply a similar method for $VW'$ appearing below:

$\frac{\sigma_2}{\sigma_1} \times VW',$

with

$$VW'=C_1Z_1^2-C_2Z_2^2$$

in which $V\sim N(\mu_1,\sigma_1^2), W'= \frac{\sigma_1}{\sigma_2} W\sim N(\frac{\sigma_1}{\sigma_2}\mu_1,\sigma_1^2)$, and $\sigma'_{12}=\frac{\sigma_1}{\sigma_2}\sigma_{12}$.

Remark: Let $U$ and $V$ have a bivaraite normal distribution and have the same variance, then $U+V$ and $U-V$ are independent (the means can be different and they can be dependent); see here for a proof.