# chi squared distribution of independent normal distributions that are not standard normal

I've been working on the following problem. I'm a bit confused about some of the specifics of how to arrive at the correct answer. I hope someone here could point me in the right direction:

A dart is tossed at a target with a Cartesian $(X_1,X_2)$ coordinate system centered in the middle of the bull's-eye. Assume $X_1$, the horizontal miss distance, and $X_2$, the vertical miss distance are independent, normal random variables, each with $\mu=0$, $\sigma^2=2$. The radial miss distance then is $R=\sqrt{X_1^2 + X_2^2}$, evaluate:

(a) $P(R>2)$

(b) The radius, $r$, of the circle centered at $(0,0)$, which has probability $0.5$ of containing $(X_1,X_2)$.

This is what I have so far:

Since $X_1$ and $X_2$ are normal variables, their square will follow the chi-squared distribution for random variables. For the "standard normal" random variable $Z$ (i.e. $\mu=0$ and $\sigma^2=1$), the random variable $Y = Z^2$ would be:

$P(Y<t) = P(Z^2<t) = P(-\sqrt{t} < Z < \sqrt{t}) = 2N(\sqrt{t})-1$

where $N(z) = \int_{-\infty}^z{\frac{1}{\sqrt{2\pi}}}e^{-t^2/2}dt$ is the cumulative distribution function for the standard normal distribution

however, these distributions are not "standard normal", since the variance ($\sigma^2$) is equal to 2, not 1.

For normal distribution functions, the cumulative distribution function is:

$F_X(x) = N\left(\frac{x-\mu}{\sigma}\right) = \int_{-\infty}^{\frac{x-\mu}{\sigma}}{\frac{1}{\sqrt{2\pi}}}e^{-t^2/2}dt$

So, the square of the normal distribution function of the random variable $W=X^2$ should follow as:

$P(W<t) = P(X^2<t) = P(-\sqrt{t}<X<\sqrt{t}) = 2N\left(\frac{\sqrt{t}-\mu}{\sigma}\right)-1 = F_W(t)$

and the density function for these random variables would be:

$f_W(t) = \frac{d}{dt}F_W(t) = \frac{n\left(\frac{\sqrt{t}-\mu}{\sigma}\right)}{\sigma\sqrt{t}} = \frac{1}{\sigma\sqrt{2\pi t}}e^{-(\sqrt{t}-\mu)^2/2\sigma^2}$

and, since the average ($\mu$) = 0, this simplifies to:

$f_W(t) = \frac{1}{\sigma\sqrt{2\pi t}}e^{-t/2\sigma^2}$

which is, of course, a gamma distribution with parameters $\lambda = \frac{1}{2\sigma^2}$ and $n = \frac{1}{2}$

Now, up to this point, I'm fairly confident, it's the next part that I'm a bit confused by.

Now, we have the random variable $R = \sqrt{X_1^2+X_2^2}$, the probability that the radius hit is greater than 2 is:

$P(R>2) = 1-P(R<2) = 1-P(\sqrt{X_1^2+X_2^2} <2) = 1 - P(X_1^2+X_2^2 < 4)$

Now, if we define a random variable $K = X_1^2 + X_2^2$, this will be a sum of independent chi-squared variables, therefore it will be a gamma distribution with $\lambda = \frac{1}{2\sigma^2}$ and $n = 2/2 = 1$:

$f_K(t) = \frac{1}{2\sigma^2} e^{-t/2\sigma^2}$

which is the Erlang probability ($P(K<t) = F_K(t) = 1-\sum_{k=0}^{r-1}{\frac{(\lambda t)^k}{k!}e^{-\lambda t}}$) law with r=1, therefore:

$P(K>t) = e^{-t/2\sigma^2} = e^{-t/4}$

so, for part (a):

$P(R>2) = P(X_1^2+X_2^2>4) = P(K>4) = e^{-4/4} = e^{-1}$

which is the correct answer according to the solutions manual

Part (b) asks to find the radius when the probability is 0.5 of hitting the target, therefore:

$P(R<r) = 1-e^{-r/4} = 0.5$

$e^{-r/4} = 0.5$

$\frac{-r}{4} = -ln(2)$

$r = ln(16)$

$r = 2.7725887...$

but since this is actually the radius squared, the value should be:

$radius = \sqrt{2.7725887} = 1.66511$

However, this is apparently incorrect. The correct answer for part (b) should be: 1.1774. This is the answer if you divide 2.7725887 by 2 before taking the square-root. I am very confused by why I've arrived at the wrong answer for part (b). If anyone could help, I would be very grateful!

I suspect you did too much work.

Let's make some observations. $$R = \sqrt{X_1^2+X_2^2} = \sqrt{\left(\sqrt 2\frac{X_1}{\sqrt 2}\right)^2+\left(\sqrt 2\frac{X_2}{\sqrt 2}\right)^2} = \sqrt 2\sqrt{\mathsf X_1^2+\mathsf X_2^2} = \sqrt 2\mathsf R$$ where $$\frac{X_1}{\sqrt 2} =\mathsf X_1\overset d= \mathsf X_2\sim N(0,1)$$ and hence $\mathsf R$ follows a standard Rayleigh distribution. So depending on the material covered, you might have overlooked this fact, or this is just another way to verify your answers.

For a) \begin{align*} P(R>2)&= P(\sqrt 2 \mathsf R >2)\\ &=1-P(\mathsf R \leq 2/\sqrt 2)\\ &=1-\left(1-\exp\left\{-\frac{1}{2}\left(\frac{2}{\sqrt 2}\right)^2\right\}\right)\tag 1\\ &=e^{-1}\\ &=0.3678794 \end{align*} where in $(1)$ I used the well known cdf of a Rayleigh.

Now for part b)

\begin{align*} P(R<r)&= \frac{1}{2}\\ P(\sqrt 2 \mathsf R <r) &= \frac 12\\ P\left(\mathsf R<\frac{r}{\sqrt 2}\right) &=\frac 12\\ 1-\exp\left\{-\frac{1}{2}\left(\frac{r}{\sqrt 2}\right)^2\right\} &=\frac{1}{2}\tag2\\ \exp\left\{-\frac{1}{2}\left(\frac{r}{\sqrt 2}\right)^2\right\}&=\frac{1}{2}\\ -\frac{1}{2}\frac{r^2}{2} &= -\log 2\\ r^2 &= 4\log 2 \end{align*} where in $(2)$ I used the well known cdf of a Rayleigh, and this gives that $r = 1.665109$.

• Thank you for the answer, however part (b) still has the technically "incorrect" answer. Although, there is a possibility that the solutions in the back of the book are wrong (they have been before) – lstbl Feb 28 '16 at 14:48
• @lstbl Let me put it bluntly. They are wrong. The three of us agree, and we used different methods. I highly doubt the solution is 'correct'. – Em. Feb 28 '16 at 14:55
• @probablyme Nice solution, but perhaps you have a typo in part (a). The Rayleigh CDF is $1-e^{-x^2/2}$, therefore $1-(1-e^{-x^2/2})=e^{-x^2/2}=1/e$, in agreement with both myself, lstbl (and the book, which seems to be the least reliable of the currently available sources...;-) – Pierpaolo Vivo Feb 28 '16 at 15:28
• ok (the numerical 0.6321206 is also to be edited ;-)) ) – Pierpaolo Vivo Feb 28 '16 at 15:33
• @PierpaoloVivo Wow, sorry I was not paying attention. Thanks a lot. – Em. Feb 28 '16 at 15:39

Let us first compute in the quickest way the PDF of $R$ using Dirac delta. The joint PDF of $X_1$ and $X_2$ is $$f_{X_1,X_2}(x_1,x_2)=\frac{e^{-x_1^2/4}}{\sqrt{4\pi}}\frac{e^{-x_2^2/4}}{\sqrt{4\pi}}\ ,$$ therefore the PDF of $R$ is $$f_R(r)=\int_{-\infty}^\infty dx\int_{-\infty}^\infty dy\frac{e^{-x_1^2/4}}{\sqrt{4\pi}}\frac{e^{-x_2^2/4}}{\sqrt{4\pi}}\delta\left(r-\sqrt{x^2+y^2}\right)\ ,$$ which is easily handled in polar coordinates $x=\hat{r}\sin\theta$ and $y=\hat{r}\cos\theta$, yielding $$f_R(r)=\frac{2\pi}{4\pi}\int_0^\infty d\hat{r}\ \hat{r} e^{-\hat{r}^2/4}\delta(r-\hat{r})=\frac{1}{2}r e^{-r^2/4}\ ,$$ which is correctly normalized for $r\in (0,\infty)$. To compute part (A), we just need to integrate this PDF for $r>2$: $$P(R>2)=\int_2^\infty dr f_R(r)=\int_2^\infty dr\frac{1}{2}r e^{-r^2/4}=1/\mathrm{e}\approx 0.3678....$$ For part (B), we just need to integrate the radial PDF $f_R(r^\prime)$ up to an unspecified point $r$, and then solve a simple equation $$P(0<R<r)=\int_0^r\ dr^\prime f_R(r^\prime)=1-e^{-\frac{r^2}{4}}\ ,$$ and then set this value equal to $0.5$: $$1-e^{-\frac{r^2}{4}}=1/2\Rightarrow e^{-r^2/4}=1/2\Rightarrow e^{-r^2/4}=e^{\ln(1/2)}\Rightarrow r=2\sqrt{\ln 2}\approx 1.66511...\ ,$$ as you've done. So your solution manual must have a typo.

• Could you give a reference to some background about using the dirac delta function to arrive at your PDF? I've never seen this usage. – lstbl Feb 28 '16 at 14:51
• @lstbl you can start here pvamu.edu/Include/Math/AAM/Vol3_No1/… and www.ncc.org.in/download.php?f=NCC2008/2008_A5_1.pdf ... once you go Dirac, you never go back! – Pierpaolo Vivo Feb 28 '16 at 15:20