# Problem with chi-squared test

I'm attempting to solve an old exam question and am running into some difficulty. Would appreciate if anyone could tell me if my approach is correct and tell me how to get to next step. Here is the question.

"A dart player throws a dart on a target centered on the bulls eye. The impact coordinates $(X,Y)$ of the dart are assumed to be independent normal in either direction with mean 0 and variance $\sigma^2$. Assume $0<\alpha<1.$ Find $r$(in terms of $\sigma$ and $\alpha$) such that

$$Pr(\sqrt{X^2+Y^2}\leq r)=1-\alpha$$"

So here is what I have so far. If $X$ and $Y$ are independent normal in either direction with mean 0 and variance $\sigma^2$ then $\frac{X}{\sigma}$ and $\frac{Y}{\sigma}$ are standard normal. Then $\frac{X}{\sigma}^2+\frac{Y}{\sigma}^2$ is chi-squared with two degrees of freedom. But how I then use this to get a relation involving $\sigma$ and $\alpha$ I'm not sure. I know the pdf of chi-squared with two degrees of freedom is $$\frac{e^{\frac{-x}{2}}}{2}$$. But what is $x$ here? and I dont want $$Pr(\frac{X}{\sigma}^2+\frac{Y}{\sigma}^2\leq r)=1-\alpha$$ I want $$Pr(\sqrt{X^2+Y^2}\leq r)=1-\alpha$$ but it seems only way to relate the quantity $\sqrt{X^2+Y^2}$ is the make it into chi-squared. Also this is part (c) of a question and (a) and (b) were about chi-squared. We don't have access to statistical tables in this exam either.

Can anyone shed some light onto the solution?

If $X$ and $Y$ are independent $\mathcal N(0,\sigma^2)$ random variables, then $R=\sqrt{X^2+Y^2}$ has what is called a Rayleigh distribution having density function $f(x)=\frac{x}{\sigma^2}e^{-x^2/(2\sigma^2)}\mathbf1_{x>0}$.

Distribution function of $R$ is given by $F(x)=\frac{\mathrm{d}}{\mathrm{d}x}f(x)=1-e^{-x^2/(2\sigma^2)}\mathbf1_{x>0}$.

So just solve for $r$ from the equation $F(r)=1-\alpha$.

Note that $\frac{X^2}{\sigma^2}$ and $\frac{Y^2}{\sigma^2}$ are independent Chi-square variables with 1 degree of freedom, so that $Z=\frac{X^2}{\sigma^2}+\frac{Y^2}{\sigma^2}\sim\chi^2_{(2)}$. Now you can calculate the CDF of $Z$, call that $F_Z(\cdot)$.

You would have $F_Z(z)=\Pr(\frac{X^2}{\sigma^2}+\frac{Y^2}{\sigma^2}\leqslant z)=\Pr(X^2+Y^2\leqslant \sigma^2z)$

$\qquad\qquad\qquad\qquad\quad=\Pr(\sqrt{X^2+Y^2}\leqslant \sigma\sqrt{z})$.

In this function of $z$, substitute $\sigma\sqrt{z}=r$ to get a function of $r$. Now I think you can proceed.

• thanks but Rayleigh distribution was not covered in my course so I don't think we are expected to do it using something not covered in lectures. Is there any way to do it in terms of chi-squared distribution? – user24907 Apr 1 '18 at 16:49
• @user24907 If you calculate $P(X^2+Y^2\le \sigma r^2)$, then sure you can. – StubbornAtom Apr 1 '18 at 16:52
• but $X^2+Y^2$ is not chi-squared. $\frac{X}{\sigma}^2+\frac{Y}{\sigma}^2$ is and even then I dont know how to interpret the expression I get. – user24907 Apr 1 '18 at 17:00
• @user24907 See my edit. – StubbornAtom Apr 1 '18 at 17:18
• @user24907 Note that it is possible to find the CDF of $\chi^2_{(2)}$ in closed form. – StubbornAtom Apr 1 '18 at 17:22