MHB Calculate Probability: Diff $Y-X \leq 1$

[evinda]

Hello! (Wave)

Suppose that $X$ has the uniform distribution on the interval $[0,2]$ and $Y$ has the uniform distribution on the interval $[2,4]$. If $X,Y$ are independent, I want to find the probability that the difference $Y-X$ is $\leq 1$.

I have thought the following.The density function of $X$ is

$$p_1(x)=\left\{\begin{matrix}
\frac{1}{2} &, 0 \leq x \leq 2 \\
0 & , \text{ otherwise}
\end{matrix}\right.$$

while the density function of $Y$ is

$$p_2(x)=\left\{\begin{matrix}
\frac{1}{2} &, 2 \leq x \leq 4 \\
0 & , \text{ otherwise}
\end{matrix}\right.$$How can we find the probability that the difference $Y-X$ is $\leq 1$ ? Do we use the above density functions? (Thinking)

 

[I like Serena]

Hey evinda! (Smile)

Yep, we use those density functions.

$X$ and $Y$ have a so called joint probability distribution $f_{X,Y}(x,y)$.
If means that:
$$P(Y-X\le 1) = \iint_{y-x\le 1} f_{X,Y}(x,y)\,dx\,dy$$
And since they are independent, we have:
$$f_{X,Y}(x,y) = f_X(x)f_Y(y) = p_1(x)p_2(y)$$
(Thinking)

 

[I like Serena]

Just for fun, here's the corresponding graph.
[DESMOS]advanced: {"version":5,"graph":{"squareAxes":false,"viewport":{"xmin":-0.5,"ymin":-0.5,"xmax":6.5,"ymax":6.5}},"expressions":{"list":[{"id":"graph1","type":"expression","latex":"\\max\\left(\\left|x-1\\right|,\\left|y-3\\right|\\right)\\le1","color":"#60a0e3"},{"id":"2","type":"expression","latex":"y-x\\le1\\left\\{\\max\\left(\\left|x-1\\right|,\\left|y-3\\right|\\right)\\le1\\right\\}","color":"#c74440"},{"id":"3","type":"expression","color":"#fa7e19"}]}}[/DESMOS]

The blue area corresponds to the part where the joint probability density is non-zero.
And the red area corresponds to the part where $Y-X \le 1$. (Thinking)

 

[evinda]

Nice, thank you! (Smirk)