Dropping rocks with the x- and y-coordinates being normal

[Eclair_de_XII]

Homework Statement

"A helicopter drops rocks onto a desert until it hits a marked bullseye. With respect to Cartesian coordinates whose origins is at the bullseye, both the x- and y-coordinates are normally distributed. Denote the r.v.'s taking on values along these axes as $X,Y$, respectively, and note that they are independent. Also note that $X,Y$ have distributions $N(0,\sigma^2)$. Show that the expectation of the distance between the landing point and the target is $\sigma \sqrt{\frac {\pi}{2}}$. Find also the variance."

Homework Equations

Hints given: "Find $P(X^2+Y^2 \leq r^2)$ and use that distribution to find $P(\sqrt{X^2+Y^2}\leq r)$, $Var(\sqrt{X^2+Y^2})$ and $E(\sqrt{X^2+Y^2})$.

If $X$ has distribution $N(\mu,\sigma^2)$, then $f_X(x)=\frac {1}{\sqrt{2 \pi} \sigma} e^{-\frac {1}{2}(\frac {x-\mu}{\sigma})^2}$.

For two independent r.v.'s $X,Y$, $f_{X,Y}(x,y)=f_X(x)f_Y(y)$.

The Attempt at a Solution

So far, what I have is:

$P(X^2+Y^2 \leq r^2)=2P(Y\leq \sqrt{r^2-x^2}|X=x)P(X=x)=2\int_{-r}^{r} \int_0^{\sqrt{r^2-x^2}} f_Y(y)f_X(x)\,dy\,dx\\ =4\int_0^r \int_0^{\sqrt{r^2-x^2}} \frac{1}{2\pi \sigma^2} e^{-\frac{1}{2\sigma^2}y^2}e^{-\frac{1}{2\sigma^2}x^2}\,dy \,dx=\frac{2}{\pi \sigma^2} \int_0^r \int_0^{\sqrt{r^2-x^2}}e^{-\frac{1}{2\sigma^2}(y^2+x^2)}\,dy \,dx$

This integral isn't something I learned how to do yet, so I'm thinking that there's another way I'm supposed to do this problem.

 

[RPinPA]

It will be easier to work out in polar form. You are integrating the interior of a circle of radius r. Express that shape in terms of $r$ and $\theta$, and also express your joint probability in terms of $r$ and $\theta$.

What is the surface element $dy dx$ expressed in polar coordinates?

 

[vela]

So far, what I have is:

$P(X^2+Y^2 \leq r^2)=2P(Y\leq \sqrt{r^2-x^2}|X=x)P(X=x)$

The righthand side isn't correct. For one thing, you meant $0 \le Y \le \sqrt{r^2-x^2}$. Also, you somehow need to incorporate the fact that X runs between -r and r. It's probably easier just to go straight to the integral:
$$P(X^2+Y^2 \leq r^2)= \int_{-r}^{r} \int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} f_Y(y) f_X(x)\,dy\,dx = \frac{1}{2\pi \sigma^2} \int_{-r}^{r} \int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} e^{-\frac{x^2+y^2}{2\sigma^2}}\,dy\,dx$$ Then as RPinPA suggested, change variables to polar coordinates. You'll be able to evaluate the resulting integral.

 

[Eclair_de_XII]

$P(X^2+Y^2 \leq r^2)=\frac{1}{2\pi \sigma^2} \int_{-r}^{r} \int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} e^{-\frac{x^2+y^2}{2\sigma^2}}\,dy\,dx=\frac{1}{2\pi \sigma^2} \int_0^{2\pi} \int_0^r se^{-\frac{s^2}{2\sigma^2}}\,ds\,d\theta$

Let $u=\frac{s^2}{2\sigma^2}$, $\sigma^2du=s\,ds$, and $r\mapsto \frac{r^2}{2\sigma^2}$, $0\mapsto 0$. So we have:

$P(X^2+Y^2 \leq r^2)=\frac{1}{2\pi} \int_0^{2\pi}\int_0^{\frac{r^2}{2\sigma^2}}e^{-u}\,du\,d\theta=e^{-u}|_{\frac{r^2}{2\sigma^2}}^0=1-e^{-\frac{r^2}{2\sigma^2}}$

Now I'm guessing I transform this into $P(0\leq \sqrt{X^2+Y^2}\leq r)$? Or is it just: $\frac{1}{2}P(X^2+Y^2 \leq r^2)$?

 

[RPinPA]

$P(X^2+Y^2 \leq r^2)=\frac{1}{2\pi \sigma^2} \int_{-r}^{r} \int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} e^{-\frac{x^2+y^2}{2\sigma^2}}\,dy\,dx=\frac{1}{2\pi \sigma^2} \int_0^{2\pi} \int_0^r se^{-\frac{s^2}{2\sigma^2}}\,ds\,d\theta$

That looks good. This is a result that I'm very familiar with from on the job (the application being white noise in communication signals): If you have a vector whose $x$ and $y$ coordinates are independent identically distributed normal random variables, then the magnitude follows a Rayleigh distribution (that's the density you have under the integral sign) and the phase angle is uniformly distributed.

You've just proved that well-known result.

Let $u=\frac{s^2}{2\sigma^2}$, $\sigma^2du=s\,ds$, and $r\mapsto \frac{r^2}{2\sigma^2}$, $0\mapsto 0$. So we have:

$P(X^2+Y^2 \leq r^2)=\frac{1}{2\pi} \int_0^{2\pi}\int_0^{\frac{r^2}{2\sigma^2}}e^{-u}\,du\,d\theta=e^{-u}|_{\frac{r^2}{2\sigma^2}}^0=1-e^{-\frac{r^2}{2\sigma^2}}$

Now I'm guessing I transform this into $P(\sqrt{X^2+Y^2}\leq r)$? Or is it just: $\frac{1}{2}P(X^2+Y^2 \leq r^2)$?

How are those two events related? That is, what set of points meets the condition $X^2 + Y^2 \leq r^2$ and what set of points meets the condition $\sqrt{X^2 + Y^2} \leq r$?

 

[Eclair_de_XII]

It's the closed circle of radius $r$ centered at the origin.

 

[RPinPA]

It's the closed circle of radius $r$ centered at the origin.

Which of my two questions is that the answer to?

1. What set of points meets the condition $X^2 + Y^2 \leq r^2$?
2. What set of points meets the condition $\sqrt{X^2 + Y^2} \leq r$?

 

[Eclair_de_XII]

The first one. The second one is a semi-circle above the x-axis, assuming it's bounded from below by $0$.

 

[RPinPA]

The first one. The second one is a semi-circle above the x-axis, assuming it's bounded from below by $0$.

Are you sure? So points with y negative are not included? Let's check. Suppose x = 1, y = -1 and r = 2. What is $\sqrt{x^2 + y^2}$? Is it $\leq$ r?

 

[Eclair_de_XII]

Yeah, I think they're included. Basically, any $(x,y)\in\mathbb{R}^2$ s.t. $x^2+y^2\leq r$. I'm guessing it's the same event, then?

 

[RPinPA]

Yeah, I think they're included. Basically, any $(x,y)\in\mathbb{R}^2$ s.t. $x^2+y^2\leq r$. I'm guessing it's the same event, then?

Correct. Those are the same event. No need to guess, it is a fact that those are the same probability.

 

[Eclair_de_XII]

Something's off, here, though. If the r.v. $\sqrt{X^2+Y^2}$ is distributed as $exp(\frac{1}{2\sigma^2})$, then its mean should be $\mu=2\sigma^2\neq \sigma\sqrt{\frac{\pi}{2}}$.

 

[vela]

$r$ doesn't follow an exponential distribution. Why do you think it does?

 

[Eclair_de_XII]

It just seemed like that $r^2$ had the form of an exponential distribution. In any case, should I try to find the distribution of $r$, now?

 

[haruspex]

It just seemed like that $r^2$ had the form of an exponential distribution.

It does, but that does not mean the distribution of r is exponential.

 

[Eclair_de_XII]

Okay, so I turned in the assignment in already, but I'll just share what I have done:

Let $R=\sqrt{X^2+Y^2}$. Then $F_R(r)=1-e^{-\frac{1}{2}(\frac{r}{\sigma})^2}$

So $f_R(r)=\frac{r}{\sigma^2}e^{-\frac{1}{2}(\frac{r}{\sigma})^2}$

$E(R)=\int_0^\infty r\frac{r}{\sigma^2}e^{-\frac{1}{2}(\frac{r}{\sigma})^2}\,dr$

Let $u =r$, $du=dr$, and $dv=\frac{r}{\sigma^2}e^{-\frac{1}{2}(\frac{r}{\sigma})^2}\,dr$, $v=-e^{-\frac{1}{2}(\frac{r}{\sigma})^2}$

So... $E(R)=(-re^{-\frac{1}{2}(\frac{r}{\sigma})^2})|_0^\infty+\int_0^\infty -e^{-\frac{1}{2}(\frac{r}{\sigma})^2}dr$

Let $s=\frac{r}{\sigma\sqrt{2}}$, $\sigma \sqrt{2}\,ds=dr$

And we have: $E(R)=\sigma \sqrt{2}\int_0^\infty e^{-s^2}\,ds=\sigma \sqrt{2}\frac{1}{2}\int_{\mathbb{R}} e^{-s^2}\,ds=\sigma \frac{1}{\sqrt{2}}\sqrt{\pi}=\sigma\sqrt{\frac{\pi}{2}}$.

Then $E(R^2)=\int_0^\infty \frac{r^3}{\sigma^2}e^{-\frac{1}{2}(\frac{r}{\sigma})^2}\,dr$. Let $t=\frac{r}{\sigma}$, $\frac{dt}{\sigma}=\frac{dr}{\sigma^2}$, and so: $E(R^2)=\sigma^2 \int_0^\infty t^3e^{-\frac{1}{2}t^2}$ and $Var(R)=\frac{\sigma^2}{2}(E(T^3)-\pi)$ where $T$ has the standard normal distribution.