Question

Suppose you are to throw a dart at a circular dart board with radius 2 inches. Let (X, Y ) denote the point that you hit on the board (you can assume the board is centered at the origin (0, 0), and that the dart hits somewhere on the board uniformly at random). (a) Find the joint PDF of X and Y . (b) Find the marginal PDFs of X and Y . (c) Find the conditional PDFs fX|Y and fY |X. (d) Suppose that the “bulls eye” on the target consists of a small circle centered at the origin with radius 0.25. Explain how you would use one of these PDFs to compute the probability of a dart hitting the bulls eye. Find this probability an even easier way. (e) What is the probability that X > 1? (f) If you know that X = 1, what is the PDF of Y conditioned on this fact? What’s the probability that Y > 0.3 conditioned on this fact?

Answer 1

a. All points on the board are equally likely to be hit with a probability of 1/(area of board), or

`f_(X,Y)(x,y)=\begin{cases}\frac1{4\pi}&\text{for }x^2+y^2\le4\\\\0&\text{otherwise}\end{cases}`

b. To find the marginal distribution of
`X`, integrate the joint distribution with respect to
`y`, and vice versa. We can take advantage of symmetry here to compute the integral:

`\displaystyle\int_y f_(X,Y)(x,y)\,\mathrm dy=2\int_0^(√(4-x^2))(\mathrm dy)/(4\pi)=(√(4-x^2))/(2\pi)`

`f_X(x)=\begin{cases}(√(4-x^2))/(2\pi)&\text{for }-2\le x\le2\\\\0&\text{otherwise}\end{cases}`

and by the same computation you would find that

`f_Y(y)=\begin{cases}(√(4-y^2))/(2\pi)&\text{for }-2\le y\le2\\\\0&\text{otherwise}\end{cases}`

c. We get the conditional distributions by dividing the joint distributions by the respective marginal distributions:

`f_(X\mid Y=y)(x)=(f_(X,Y)(x,y))/(f_Y(y))`

`f_(X\mid Y=y)(x)=\begin{cases}\frac1{2√(4-y^2)}&\text{for }-2\le y\le2\text{ and }x^2\le4-y^2\\\\0&\text{otherwise}\end{cases}`

and similarly,

`f_(Y\mid X=x)(y)=\begin{cases}\frac1{2√(4-x^2)}&\text{for }-2\le x\le2\text{ and }y^2\le4-x^2\\\\0&\text{otherwise}\end{cases}`

d. You can compute this probability by integrating the joint distribution over a part of the circle (call it "B" for bullseye):

`\displaystyle\iint_Bf_(X,Y)(x,y)\,\mathrm dx\,\mathrm dy=\int_0^(2\pi)\int_0^(0.25)\frac r{4\pi}\,\mathrm dr\,\mathrm d\theta=\frac1{64}`

(using polar coordinates) The easier method would be to compute the area of a circle with radius 0.25 instead, then divide that by the total area of the dartboard.

`(\pi\left(\frac14\right)^2)/(\pi\cdot2^2)=\frac1{64}`

e. The event that
`X>1` is complementary to the event that
`X\le1`, so

`P(X>1)=1-P(X\le1)=1-F_X(1)`

where
`F_X(x)` is the marginal CDF for
`X`. We can compute this by integrate the marginal PDF for
`X`:

`F_X(x)=\displaystyle\int_(-\infty)^xf_X(t)\,\mathrm dt=\begin{cases}0&\text{for }x<-2\\\\\frac12+\frac1\pi\sin^(-1)\frac x2+(x√(4-x^2))/(4\pi)&\text{for }-2\le x<2\\\\1&\text{for }x\ge2\end{cases}`

Then

`P(X>1)=1-F_X(1)=\frac13-(\sqrt3)/(4\pi)\approx0.1955`

f. We found that either random variable conditioned on the other is a uniform distribution. In particular,

`f_(Y\mid X=1)(y)=\begin{cases}\frac1{2\sqrt3}&\text{for }y^2\le3\\\\0&\text{otherwise}\end{cases}`

Then

`P(Y>0.3\mid X=1)=1-P(Y\le0.3\mid X=1)=1-F_(Y\mid X=1)(0.3)`

where
`F_(Y\mid X=x)(y)` is the CDF of
`Y` conditioned on
`X=x`. This is easy to compute:

`F_(Y\mid X=1)(y)=\displaystyle\int_(-\infty)^yf_(Y\mid X=1)(t)\,\mathrm dt=\begin{cases}0&\text{for }y<-\sqrt3\\\\(y+\sqrt3)/(2\sqrt3)&\text{for }-\sqrt3\le y<\sqrt3\\\\1&\text{for }y\ge\sqrt3\end{cases}`

and we end up with

`P(Y>0.3\mid X=1)=(10-\sqrt3)/(20)\approx0.4134`