# Finding $f_Y$ such that $Z=Y\cos(X)\sim\mathcal{N}_{0,\sigma}$ for $X\sim\mathcal{U}[0,2\pi]$

I need to choose the probability distribution $f_Y(y)$ of a random variable $Y$ such that the variable $Z=Y\cos(X)$ is normally distributed with zero mean, i.e. $$f_Z(z)=\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{z^2}{2\sigma^2}}$$ for the independent and uniform distributed variable $X$, i.e. $$f_X(x)=\frac{1}{2\pi}$$ for $x\in[0,2\pi]$.

I started looking for the probability distribution of $K=\cos(X)$ and found $f_K(k)=\frac{1}{\pi\sqrt{1-k^2}}$ for $k\in[-1,1]$. Next, I wrote down the product distribution for $Y$ and $K$ (http://en.wikipedia.org/wiki/Product_distribution)

$$f_Z(z)=\int_{-\infty}^\infty{f_Y(x)f_K\left(\frac{z}{x}\right)\frac{1}{|x|}dx}$$

and put in what I have:

$$\int_{|z|<|x|}{f_Y(x)\frac{1}{\pi\sqrt{1-z^2/x^2}}\frac{1}{|x|}dx}\stackrel{!}{=}\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{z^2}{2\sigma^2}}$$

But now I´m stuck. How to find the $f_Y$ that fulfills this equation?

Great, $Y=\sqrt{-2 \ln(U)}$ with $U\sim\mathcal{U}[0,1]$ does the job for $Z\sim\mathcal{N}_{0,1}$ and thus $Y=\sqrt{-2\sigma ^2 \ln(U)}$ leads to $Z\sim\mathcal{N}_{0,\sigma}$, so:

$$f_Y(y)=\frac{y}{\sigma ^2}e^{-\frac{y^2}{2\sigma ^2}}$$

Extension to nonzero mean

Now how would one go about extending this to allow for nonzero means, i.e. $Z´=Y´\cos(X)\sim\mathcal{N}_{\mu,\sigma}$?

We can write $Z´=Z+\mu$ and thus $Y´=Y+\mu/\cos(X)$ and find for $Q=\mu/\cos(X)$ $f_Q(q)=\frac{|\mu|}{q^2\sqrt{1-\mu^2/q^2}}$ for $|q|>|\mu|$. Then, because Y and Q are independent, the probability distribution for $Y´$ should be the convolution $$f_{Y´}(y´)=\int_{-\infty}^\infty{f_Y(y´-\tau)f_Q(\tau)d\tau}$$ correct? And now i´m stuck on this :-)

$f_{Y´}(y´)$ should look like the blue curve here: Suppose $Y$ is exponentially distributed with expected value $\sigma$ so that $\Pr(Y>y)=e^{-y/\sigma}$, and let \begin{align} Z_1 & = \sqrt{-2\log_e Y}\,\cos X, \\ Z_2 & =\sqrt{-2\log_e Y}\,\sin X. \end{align} Then each of $Z_1,Z_2$ is normally distributed with expected value $0$ and variance $\sigma^2$ and $Z_1,Z_2$ are indepedent.

Postscript in response to comments: If $Z_1,Z_2\sim\text{i.i.d. }N(0,1)$ then their joint density is $$f_{(Z_1,Z_2)} (z_1,z_2) = \text{constant}\cdot e^{-z_1^2/2} e^{-z_2^2/2} = \text{constant} \cdot e^{-(z_2^2+z_2^2)/2}.$$ Since this density is constant on circles defined by $z_a^2+z_2^2=\text{radius}^2$, the angle that the random point makes with the positive $z_1$-axis is uniformly distributed between $0$ and $2\pi$. Hence its coordinates are the cosine and sine of a uniformly distributed random angle. Then the remaining question is that of the distribution of the distance of the random point from the origin.

• This is assuming one knows the answer and checking it, not finding the answer. – Did Sep 22 '14 at 14:26
• @Did : Perhaps, but what "one knows" from standard courses in my experience would include the fact that if $Z_1,Z_2\sim\text{i.i.d. }N(0,1)$ then $Z_1^2+Z_2^2\sim\chi_2^2$, and that is an exponential distribution with expected value $2$, and that the bivariate normal distribution with covariance $0$ has a circular symmetry (hence $(\cos X,\sin X)$, where $X$ is uniformly distrtibuted between $0$ and $2\pi$). ${}\qquad{}$ – Michael Hardy Sep 22 '14 at 14:31
• Not judging from what the OP shows (since, for once, we have an OP who actually shows something...). – Did Sep 22 '14 at 14:49

We have: $$\begin{eqnarray*}\mathbb{P}\left[Y=\frac{Z}{\cos(X)}\in[0,\eta]\right]&=&\frac{1}{\sqrt{2}\,\sigma\,\pi^{3/2}}\int_{0}^{\eta}\int_{-1}^{1}\frac{e^{-\frac{(ux)^2}{2\sigma^2}}}{\sqrt{1-x^2}}\,dx\,du\\&=&\frac{1}{\sigma\sqrt{2\pi}}\int_{0}^{\eta}e^{-\frac{u^2}{4\sigma^2}}\,I_0\left(\frac{u^2}{4\sigma^2}\right)\,du\end{eqnarray*}$$ where $I_0$ is a modified Bessel function of the first kind, giving:

$$f_Y(y) = \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{y^2}{4\sigma^2}}\sum_{m=0}^{+\infty}\frac{1}{m!^2}\left(\frac{y^4}{64\sigma^4}\right)^m.$$

The pdf of $Y$ in a neighbourhood of zero can be very well approximated by the pdf of a Cauchy distribution.

• Seems to me you're making this too complicated. Doesn't the standard Box-Muller transformation do it? – Michael Hardy Sep 22 '14 at 14:17
• Box-Muller or Mellin, the pdf of $Y$ is still the product of a Gaussian and a Bessel function. – Jack D'Aurizio Sep 22 '14 at 14:19