I have two independent random variables, $Y$ and $X,$ where $Y$ is a random variable with a Gaussian distribution and a zero mean.  $X$ is a random variable with a Gaussian distribution and a zero mean.  I then create a new random variable:

$$Z = Y+Y*X$$

Is there any way to determine the PDF of $Z$ and how can I compute the correlation between $Y$ and $YX$ if it exists?

  • 1
    $\begingroup$ Are the variances of those X and Y known? Are they 1.0? $\endgroup$
    – Alex
    Sep 29, 2022 at 22:46
  • 3
    $\begingroup$ Z is the product of two Gaussian random variables Y and 1+X, so ...... $\endgroup$ Sep 29, 2022 at 22:53
  • $\begingroup$ Assuming independence and zero means: mathworld.wolfram.com/NormalProductDistribution.html $\endgroup$
    – Galen
    Sep 29, 2022 at 23:30
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    $\begingroup$ You don't need the pdf of $Z=Y+XY$ to compute the correlation of $Y$ and $XY$. Note that since $X$ and $Y$ are independent zero-mean random variables, \begin{align}\operatorname{cov}(Y,XY)&=E[Y^2X]-E[Y]E[XY]\\&=E[Y^2]E[X]-(E[Y])^2E[X]\\&=\sigma_Y^2\cdot 0 - 0\cdot 0\\&=0\end{align} and so $Y$ and $XY$ are uncorrelated random variables. $\endgroup$ Sep 30, 2022 at 1:55

1 Answer 1


Let $U = (X+Y)/\sqrt{2}$ and $V = (X-Y)/\sqrt{2}.$ Because these are linear combinations of the bivariate Normal variable $(X,Y),$ $(U,V)$ has a bivariate Normal distribution. Direct calculation of the means and covariances shows $U$ and $V$ are uncorrelated standard Normal variables, whence they are independent.

Algebra shows that

$$Z = Y + YX = ((U + \sqrt{1/2})^2 - (V + \sqrt{1/2})^2)/2,$$

a difference of two independent identically distributed terms. Each of these terms has--by definition--a noncentral $\chi^2(1,1/2)$ distribution.. Consequently, writing $\psi$ for the common characteristic function, the c.f. of $Z$ must be

$$\psi_{Z}(t) = \psi(t/2)\psi(-t/2) = \frac{\exp\left(\frac{it/4}{1-it}\right)}{\sqrt{1-it}} \frac{\exp\left(\frac{-it/4}{1+it}\right)}{\sqrt{1+it}} = \frac{1}{\sqrt{1 + t^2}} \exp\left(\frac{-t^2/2}{1+t^2}\right).$$

The density of $Z$ is the inverse Fourier transform of $\psi_{Z}.$ It can be computed as a cosine transform because $Z$ is (obviously) symmetric,

$$f_{Z}(z) = \frac{1}{2\pi} \int_{0}^{\infty} \frac{2 \cos(t z)}{\sqrt{1 + t^2}} \exp\left(\frac{-t^2/2}{1+t^2}\right)\,\mathrm{d}t.$$

100,000 simulated values of $X$ and $Y$ produced this histogram of $Z.$ On it I have plotted, in red, the values of $f_Z$ computed by numerically integrating this formula.

enter image description here

The integration could be a little dicey because the integral, as a Riemann integral, does not converge absolutely. However, it does converge and, as it turns out, the calculation is not onerous. A PC workstation can compute several thousand values a second.

Because even substantial algebraic errors in the calculation could still produce a close agreement in this plot, I simulated a million values of $Z,$ of which this next histogram is a detail near the origin where the integration becomes most difficult:

enter image description here

The agreement remains excellent.

BTW, because $X$ and $Y$ have finite variance, the covariance of $Y$ and $YX$ exists and is finite. Moreover, since $X$ and $-X$ have the same distribution,

$$\operatorname{Cov}(Y, YX) = \operatorname{Cov}(Y, Y(-X)) = -\operatorname{Cov}(Y, YX)$$

implies this covariance must be $0.$


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