probabilityuniform distribution
How can I compute this probability? I do not know what to do since it involves two random variables.
Let $X$ and $Y$ be uniform random variables on $(0,1)$. How can I compute this?
$$
P(|X-Y| < 0.25).
$$
I tried to do it using an integral
$$
\int_0^1 P(|X-y| < 0.25) \,dy
$$
but I do not know what to do next.
edit: I forgot to mention the independence of the random variables. Thanks for warning.
The desired probability corresponds to the volume of the subset of the unit cube $[0,1]^3$ that is bounded by the three planes $x+y=z$, $x+z=y$, $y+z=x$. Each of these planes chops off a tetrahedron (e.g. the one with vertices $(0,0,0)$, $(1,0,1)$, $(0,1,1)$ and $(0,0,1)$ for the plane $x+y=z$) of volume $\frac 16$. These tetrahedra are disjoint (only the biggest number can be bigger than the sum of the other two numbers), hence the volume remaining is $$1-3\cdot \frac16=\frac12.$$
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}
Best Answer
Hint: In order to compute an answer, you will have to make an assumption. One natural one is that $X$ and $Y$ are independent. Perhaps that was included in the question, and you forgot to mention it. Or perhaps it was (by mistake) left out.
Assuming independence, the pair $(X,Y)$ has uniform distribution on the square. So the joint density function is $1$ in the square, and $0$ outside.
The answer is then $$\iint_A 1\,dy\;dx,$$ where $A$ is the part of the square where $|x-y|<0.25$, or if you wish, $\le 0.25$, it makes no difference to the answer.
We can now do the integration, maybe by expressing our double integral as an iterated integral. But there is a much simpler way to solve the problem. We are integrating $1$ over $A$, so the result is the area of $A$.
Draw the region $A$ carefully. After you do that, you will not find it difficult to find its area. No integration, just basic geometry.
Remark: To see that we need some sort of assumption about $X$ and $Y$, let $X$ be uniformly distributed on $(0,1)$, and let $Y=X$. Then $X$ and $Y$ satisfy the conditions of the problem as stated, but are very much not independent. It is clear that $P(|X-Y|<0.25)=1$.