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

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 $$f_A(x)$$?

• My confusion is with the square root in place. How to deal with it and find CDF and PDF for A? Jan 13, 2021 at 10:04

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.

• So does it mean A is distributed as Chi distribution with degree of freedom 2 or Rayleigh distribution? Jan 13, 2021 at 10:17
• @TuhinDutta I have edited my answer. Jan 13, 2021 at 11:39
• @Rama Murthy sir, thank you. Jan 13, 2021 at 18:41

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

• So does it mean A is distributed as Chi distribution with degree of freedom 2 or Rayleigh distribution? Jan 13, 2021 at 10:17
• @TuhinDutta yes, you are right. Jan 13, 2021 at 14:50
• sir, thank you. Jan 13, 2021 at 18:43

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.

• sir, thank you. Jan 13, 2021 at 18:42