Given $X$ and $Y$ are independent N(0,1) random variables and $Z = \sqrt{X^2+Y^2}$ from the marginal pdf of $Z$

Question

Let $X$ and $Y$ be independent $N(0; 1)$ random variables. Let $Z = \sqrt{X^2+Y^2}$.

(a) Derive the marginal pdf of $Z$ and then using the marginal pdf to compute ${\rm E}[Z^2]$

(b) Can you propose a different way other than that in (a) to compute ${\rm E}[Z^2]$

(c) Compute ${\rm E}[Z]$.

This is the whole question that I was asked. I can do (c) but I don't know how to find the marginal pdf of $Z$ with the given information and can't seem to find any formulas. Any help would be appreciated.

Answer 1

If $X_1, X_2, \ldots, X_n$ are independent and identically distributed standard normal (mean 0, variance 1) random variables, then $$S = X_1^2 + X_2^2 + \cdots + X_n^2$$ is a chi-squared random variable with $n$ degrees of freedom. The probability density function of $S$ is given by $$f_S(s) = \frac{s^{n/2-1} e^{-s/2}}{\Gamma(n/2)2^{n/2}}, \quad s > 0.$$ In your case, $n = 2$, so $$f_S(s) = \frac{1}{2} e^{-s/2}, \quad s > 0.$$ This also happens to be an exponential distribution with mean $2$. Now, we are given $Z = \sqrt{S}$. So to find the density for $Z$, we consider the cumulative distribution function: $$F_Z(z) = \Pr[Z \le z] = \Pr[\sqrt{S} \le z] = \Pr[S \le z^2].$$ Calculate this last probability and then take the derivative to obtain the density for $Z$. Then compute an appropriate integration. Can you do the rest?

Answer 2

Using heropup's answer, we have $$ f_S(s) = \frac{1}{2} e^{-\frac{1}{2}s}, \quad s > 0 $$ and $$ F_Z(z) = \Pr[S \le z^2]. $$ Now, we determine the CDF of $S$. $$ \begin{align} F_S(s)&=\int_0^sf_S(t)\ dt\\ &=\int_0^s\frac{1}{2} e^{-\frac{1}{2}t}\ dt,\quad\text{let }\ u=-\frac{1}{2}t\quad\Rightarrow\quad dt=-2\ du\\ &=1-e^{-\frac{1}{2}s}, \end{align} $$ then $$ F_Z(z)=F_S(z^2)=1-e^{-\frac{1}{2}z^2} $$ and $$ f_Z(z)=\frac{d}{dz}F_Z(z)=\frac{d}{dz}\left(1-e^{-\frac{1}{2}z^2}\right)=z\ e^{-\frac{1}{2}z^2}. $$ Hence $$ \text{E}\left[Z^2\right]=\int_0^\infty z^2f_Z(z)\ dz=\int_0^\infty z^3e^{-\frac{1}{2}z^2}\ dz\tag1 $$ and $$ \text{E}\left[Z\right]=\int_0^\infty zf_Z(z)\ dz=\int_0^\infty z^2e^{-\frac{1}{2}z^2}\ dz.\tag2 $$ $(1)$ and $(2)$ are Gaussian integrals, where $$ \int_0^\infty z^n e^{-az^2}\ dz = \left\{ \begin{array}{l l} \frac{(n-1)!!}{2\cdot(2a)^\frac{n}{2}}\sqrt{\frac{\pi}{a}} & \quad \text{for $n$ even}\\ \\ \frac{\left(\frac{1}{2}(n-1)\right)!}{2\cdot(a)^\frac{n+1}{2}} & \quad \text{for $n$ odd.} \end{array} \right.\tag3 $$ About double factorial $(!!)$, you may refer to this link. Thus, using $(3)$, we obtain $$ \text{E}\left[Z^2\right]=\frac{\left(\frac{1}{2}(3-1)\right)!}{2\cdot\left(\frac{1}{2}\right)^\frac{3+1}{2}}=2 $$ and $$ \text{E}\left[Z\right]=\frac{(2-1)!!}{2\cdot\left(2\cdot\frac{1}{2}\right)^\frac{2}{2}}\sqrt{\frac{\pi}{\left(\frac{1}{2}\right)}} =\sqrt{\frac{\pi}{2}}. $$ For a different way to compute $\text{E}\left[Z^2\right]$, you may refer to this link. I hope this help.

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Answer 3

Hint: You may want to look at Rayleigh Distribution.