Probability : find $P(|X-y|<1/2)$ for a uniform distribution of $x,y$

Question

The question : Two guys pick a random point between $[0,2]$ (uniform distribution) . find the probability that the distance between the points is < $0.5$ $$X - \text{first point} $$ $$Y - \text{second point}$$ so because they are random $$f(x,y)=f(x)*f(y)=0.25$$

So I wrote $$P(|X-Y|<0.5) = P(-0.5<x-y<0.5)= P(x-y<0.5)-p(x-y<-0.5) =P(x<y+0.5)-p(x<y-0.5)$$

using the integrals

$$\int_0^2 \int_0^{y+0.5} 0.25 dxdy-\int_0^2 \int_0^{y-0.5}0.25 dxdy $$

I get $0.5$ and the answer is $7/16$ according to my professor.

What am I'm doing wrong ? I can draw the graph and see the limits but I think it's faster like that.

Thanks

Answer 1

Well your domain of integration is wrong. For example, $y+0.5$ goes beyond $2$.

It is always advisable to sketch a diagram.

A quick way to solve this question is simply calculate the corresponding area multiplied by the density.

$$\frac14 \left( 4 - 2 \times \frac12 \left( \frac32 \right)^2 \right) =1-\frac{9}{16}=\frac7{16}$$

Edit: \begin{align} &\int_0^2 \int_0^{y+0.5} f(x,y) \,dx \, dy - \int_0^2 \int_0^{y-0.5}f(x,y) \,dx \, dy \\ &=\int_0^2 \int_0^{y+0.5} \frac14 \mathbb{1}_{0 \leq x \leq 2, 0 \leq y \leq 2 } \,dx \, dy - \int_0^2 \int_0^{y-0.5}\frac14 \mathbb{1}_{0 \leq x \leq 2, 0 \leq y \leq 2 } \,dx \, dy \\ \\&=\int_0^2 \int_0^{\min(2,y+0.5)} \frac14\,dx \, dy-\int_{\frac12}^2 \int_0^{y-0.5} \frac14 \,dx\,dy \end{align}

Answer 2

You're wrong on your use of the probability formula. Your double inequality stands for intersection of two sets! Define $A=\lvert −0.5<x−y\lvert$, $B=\lvert x−y<0.5\lvert$, now $P(AB)=P(A)+P(B)-P(AUB)$ and you can do the rest!

Answer 3

I confirmed your professor's answer. Just write $$P(|X-Y|<\tfrac12) =\int_{-\infty}^{\infty}P(|X-y|<\tfrac12|Y=y)f_Y(y)\;dy$$ $$= \int_{-\infty}^{\infty}P(-\tfrac12<X-y<\tfrac12|Y=y)f_Y(y)\;dy$$ $$= \int_{-\infty}^{\infty}P(y-\tfrac12<X<y+\tfrac12|Y=y)f_Y(y)\;dy$$ $$= \int_{-\infty}^{\infty}[F_X(y+\tfrac12)-F_X(y-\tfrac12)]f_Y(y)\;dy\tag{$*$}$$ You know that the PDF for $X$ is $$f_X(x)=\begin{cases}0,&x\leq 0\\ \tfrac12,&0<x<2\\ 0,&x\geq 2\end{cases}$$ and the CDF for $X$ is therefore $$F_X(x)=\begin{cases}0,&x\leq 0\\ \tfrac x2,&0<x<2\\1,&x\geq 2\end{cases}$$ (and identically so for Y as well).

Now just plug in and compute the integral $(*)$.