91影视

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

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

And Z is exponential with rate $1$.

Random variables,

localid="1647259877551" $X=\sqrt{2Z}\cos U$and $Y=\sqrt{2Z}\sin U$

Density function of random vector (U,Z),

localid="1647268595855" $f_{U,Z}(u,z)=f_U\left(u\right)f_Z\left(z\right)=\frac{e^{-z}}{2\mathrm{蟺}}$

for $U鈭�(0,2\mathrm{蟺})$

And $z>0$

Now, Apply the transformation,

$g:(0,2\mathrm{蟺})脳(0,鈭�)鈫�R^2$

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

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

localid="1647268617715" $f_{X,Y}(x,y)=f_{U,Z}(u,z)路{\left|det(鈭�g(u,z\left)\right)\right|}^{-1}$

Then calculate,

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

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

Thus,

localid="1647268646133" $f_{X,Y}(x,y)=f_{U,Z}(u,z)=\frac{e^{-z}}{2\mathrm{蟺}}$

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,

localid="1647268675205" $f_{X,Y}(x,y)=\frac{e^{-z}}{2\mathrm{蟺}}=\frac{1}{2\mathrm{蟺}}{e}^{-\frac{x^2+y^2}{2}}$

This probability distribution function can be categorized as

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

Therefore,

X and Y are independent standard normal variables.