Difficult integral and joint probability (calculating something like $Pr(X \geq b-a, X \geq Y-a)$) where $X$ and $Y$ are random variables

Question

I'm trying to have some practice with joint probabilities, and I came up with a question that I'm struggling to answer.

Suppose that we have two random variables $X$ and $Y$, where $X$ is distributed uniformly in $[-\lambda, \lambda]$ and $Y$ is distributed uniformly in $[0,\phi]$. Suppose we have two constants $a$ and $b$.

How would we find this probability?

$$Pr(X \geq b-a, X \geq Y-a)$$

First, to make things a lot simpler, I assumed $X$ and $Y$ are independent. I also assumed that $\phi$ is large enough such that for any $y$ and $a, \phi > y - a$. Then, if we denote $f(x)$ and $g(y)$ as the pdfs, we can write the above as:

$$\int_{b-a}^{\lambda} \int_{y-a}^{\phi} f(x) g(y) dx dy$$

Since the second integral has a random variable in the bounds, how would we compute this? I've tried doing a convolution, but then just got lost in the math.

Any help would be appreciated!

Answer 1

$$\int_{b-a}^\lambda\int_{y-a}^\phi\frac{1}{2\lambda\phi}dxdy=\frac{1}{2\lambda\phi}\int_{b-a}^\lambda (\phi+a-y)dy=\frac{1}{2\lambda\phi}[(\phi+a)y-\tfrac{1}{2}y^2]_{b-a}^\lambda=\frac{(\lambda+a-b)(2\phi+3a-\lambda-b)}{4\lambda\phi}\cdot$$

Answer 2

Observe that: $$\{X\geq b-a,X\geq Y-a\}=\{X\geq\max(b,Y)-a\}$$ so that:$$P(X\geq b-a,X\geq Y-a)=\int P(X\geq\max(b,Y)-a)g(y)dy=$$$$\int_{-\infty}^bP(X\geq b-a)g(y)dy+\int_{b}^{\infty}P(X\geq y-a)g(y)dy$$

Work this out.