How will the plot of CDF and PDF look like for $A\ =\ \sqrt{X_1^2\ +\ X_2^2}$ where $X_1\ \sim\ N(0,1)\ and\ X_2\ \sim\ N(0,1)$.

Question

I know Chi-Square distribution. Here the expression is similar to Chi-square distribution with the degree of freedom being 2, except for the square root.

Given: $X_1\ \sim\ N(0,1)\ and\ X_2\ \sim\ N(0,1)$.

$A\ =\ \sqrt{X_1^2\ +\ X_2^2}$

How to find the CDF $F_A(x)$ and PDF $f_A(x)$?

The PDF of $\chi^2$ distribution for $\nu$ degrees of freedom is given by -

$f(\nu^2)\ =\ \frac{1}{2^{\nu / 2}.\Gamma(\nu / 2)}.e^{-\chi^2/2}.(\chi^2)^{\nu/2-1}\ \ ;\ \chi^2\ \geq\ 0$

Can I substitute 2 in place of $\nu$ to get the expression for PDF $f_A(x)$?

Answer 1

You forgot to mention independence. Without independence of $X_1$ and $X_2$ you cannot find hhe distribution of $A$.

Let $Y=X_1^{2}+X_2^{2}$. Then $f_{\sqrt Y}(z) =2zf_Y(z^{2})$ [since $P(\sqrt Y \leq z)=P(Y \leq z^{2})$ whose deirvative is $2zf_Y(z^{2})$]. Since you know the density of $Y$ you can write down the denisty of $\sqrt Y$. By definition, the disrtibution of $A$ is a chi-distribution with two degrees of freedom.

Answer 2

Take a look at Chi distribution — it's distribution of $\sqrt{X}$ for $X \sim \chi^2$.

Answer 3

If $X_1$ and $X_2$ are independent, $A$ is Rayleigh. If not, the dependency must be determined. For instance, if $X_1$ and $X_2$ are jointly normal, their correlation suffices to calculate the distribution of $A$. If not, the joint distribution is required. If the dependency is unknown, the distribution of $A$ is also unknown.