Q.6.54 - A First Course in Probability

Question

If U is uniform on $(0,2\mathrm{\pi })$ and Z, independent of U, is exponential with rate $1$, show directly (without using the results of Example $7$b) that X and Y defined by

$$\begin{gathered} X=\sqrt{2Z}\cos U \\ Y=\sqrt{2Z}\sin U \end{gathered}$$

 are independent standard normal random variables. 

Step-by-Step Solution

Answer

X and Y are independent standard normal variables with density function :

$f_{X,Y}(x,y)=\frac{1}{2\mathrm{\pi }}{e}^{-\frac{x^2+y^2}{2}}$

Step 1 : Variable :

The unknown number is represented by a variable, which is an alphabet. It represents the worth of something.

Step 2 : Explanation :

U is uniform on $\left[0,2\mathrm{\pi }\right]$.

And Z is exponential with rate $1$.

Random variables, 

$X=\sqrt{2Z}\cos U$ and $Y=\sqrt{2Z}\sin U$

Density function of random vector (U,Z),

$f_{U,Z}(u,z)=f_U\left(u\right)f_Z\left(z\right)=\frac{e^{-z}}{2\mathrm{\pi }}$

for $U\in (0,2\mathrm{\pi })$

And $z>0$

Now, Apply the transformation,

$g:(0,2\mathrm{\pi })\times (0,\infty )\to R^2$

$g(u,z)=(x,y)=\left(\sqrt{2z}\cos \right(u),\sqrt{2z}\sin (u\left)\right)$

By using theorem, the density function of random vector $(X,Y)=g(U,Z)$ as

$f_{X,Y}(x,y)=f_{U,Z}(u,z)·{\left|det(\nabla g(u,z\left)\right)\right|}^{-1}$

Then calculate,

$\nabla g(u,z)=\left[\begin{array}{cc}-\sqrt{2z}\sin \left(u\right)& \frac{\cos \left(u\right)}{\sqrt{2z}}\\ \sqrt{2z}\cos \left(u\right)& \frac{\sin \left(u\right)}{\sqrt{2z}}\end{array}\right]$

$\left|det(\nabla g(x,y\left)\right)\right|=-{\sin }^2\left(u\right)-{\cos }^2\left(u\right)=-1$

Thus,

$f_{X,Y}(x,y)=f_{U,Z}(u,z)=\frac{e^{-z}}{2\mathrm{\pi }}$

Step 3 : Explanation :

Now, write z in terms of x and y and substitute it.

But we have,

$x=\sqrt{2z}\cos \left(u\right)$

That becomes 

$x^2=2z{\cos }^2\left(u\right)$

And 

$y=\sqrt{2z}\sin \left(u\right)$

That becomes 

$y^2=2z{\sin }^2\left(u\right)$

Then sum up these to equalities.

$z=\frac{x^2+y^2}{2}$

That finally implies,

$f_{X,Y}(x,y)=\frac{e^{-z}}{2\mathrm{\pi }}=\frac{1}{2\mathrm{\pi }}{e}^{-\frac{x^2+y^2}{2}}$

This probability distribution function can be categorized as 

$\frac{1}{2\mathrm{\pi }}{e}^{-\frac{x^2+y^2}{2}}=\frac{1}{\sqrt{2\mathrm{\pi }}}{e}^{-\frac{x^2}{2}}·\frac{1}{\sqrt{2\mathrm{\pi }}}{e}^{-\frac{y^2}{2}}$.

Therefore,

X and Y are independent standard normal variables.