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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 $f_A(x)$?

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

3 Answers 3

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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.

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  • $\begingroup$ So does it mean A is distributed as Chi distribution with degree of freedom 2 or Rayleigh distribution? $\endgroup$ Jan 13, 2021 at 10:17
  • $\begingroup$ @TuhinDutta I have edited my answer. $\endgroup$ Jan 13, 2021 at 11:39
  • $\begingroup$ @Rama Murthy sir, thank you. $\endgroup$ Jan 13, 2021 at 18:41
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Take a look at Chi distribution — it's distribution of $\sqrt{X}$ for $X \sim \chi^2$.

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

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  • $\begingroup$ sir, thank you. $\endgroup$ Jan 13, 2021 at 18:42

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