Problem 3

Consider that

$U_{1} = \frac{Θ}{2 π}$ , $\frac{d U _{1}}{d Θ} = \frac{1}{2 π}$
$U_{2} = exp (- \frac{R ^{2}}{2})$ , $\frac{d U _{2}}{d R} = R exp (- \frac{R ^{2}}{2})$ Using change of variables we have:
$f_{Θ} (θ) = \frac{1}{2 π}$
$f_{R} (r) = r exp (- \frac{r ^{2}}{2})$

Note that:

$Z_{1}^{2} + Z_{2}^{2} = R^{2} (cos^{2} Θ + sin^{2} Θ) = R^{2}$ thus $R = Z_{1}^{2} + Z_{2}^{2}$
- $\frac{d R}{d Z _{1}} = Z_{1} (Z_{1}^{2} + Z_{2}^{2})^{- 1/2} = \frac{Z _{1}}{R}$
- Symmetrically $\frac{d R}{d Z _{2}} = \frac{Z _{2}}{R}$
$\frac{Z _{2}}{Z _{1}} = \frac{R s i n Θ}{R c o s Θ} = tan Θ$ thus $Θ = arctan (\frac{Z _{2}}{Z _{1}})$
- $\frac{d Θ}{d Z _{1}} = \frac{1}{1 + \frac{Z _{2}^{2}}{Z _{1}^{2}}} \cdot Z_{2} (- 1) Z_{1}^{- 2} = - \frac{Z _{2}}{Z _{1}^{2} + Z _{2}^{2}} = - \frac{Z _{2}}{R ^{2}}$
- Symmetrically $\frac{d Θ}{d Z _{2}} = \frac{Z _{1}}{Z _{1}^{2} + Z _{2}^{2}} = \frac{Z _{1}}{R ^{2}}$ Now, calculate the jacobian for a multivariate change of variables:

∣ J ∣ = \frac{\partial r}{\partial z _{1}} \frac{\partial θ}{\partial z _{1}} \frac{\partial r}{\partial z _{2}} \frac{\partial θ}{\partial z _{2}} = \frac{Z _{1}}{R} - \frac{Z _{2}}{R ^{2}} \frac{Z _{2}}{R} \frac{Z _{1}}{R ^{2}} = \frac{Z _{1}^{2}}{R ^{3}} + \frac{Z _{2}^{2}}{R ^{3}} = \frac{1}{R}

And thus the join probability being:

f_{Θ, R} (θ, r) = f_{Θ} (θ) \cdot f_{R} (r) \cdot ∣ J ∣ = \frac{1}{2 π} r exp (- \frac{r ^{2}}{2}) \cdot \frac{1}{r} = \frac{1}{2 π} exp (- \frac{z _{1}^{2} + z _{2}^{2}}{2}) = \frac{1}{2 π} exp (- \frac{z _{1}^{2}}{2}) \cdot \frac{1}{2 π} exp (- \frac{z _{2}^{2}}{2}) = f_{Z_{1}} (z_{1}) \cdot f_{Z_{2}} (z_{2})

This show both that:

$f_{Z_{1}}, f_{Z_{2}}$ is the standard normal pdf
$Z_{1}, Z_{2}$ are independent because the joint pdf is a simple product of each pdf. ∎

PK's Notes