Joint density problem. Two uniform distributions

Question

This is the problem:

An insurer estimates that Smith's time until death is uniformly distributed on the interval [0,5], and Jone's time until death also uniformly distributed on the interval [0,10]. The insurer assumes the two times of death are independent of one another. Find the probability that Smith is the first of the two to die.

The solution manual first multiplies them by one another and does this:

$$\int_0^{5}\int_s^{10}\frac{1}{50}\ dj\ ds$$

I don't get how this integral describes smith's time of death happening faster. They have a rectangle they drew with a shaded reason that I don't quite understand how describes this problem either. Can someone help me out here. I feel like I'm missing something obvious

Answer 1

Let $X$ be a random variable that denotes Smith's time until death and $0\le X\le5$. Let $Y$ be a random variable that denotes Jone's time until death and $0\le Y\le10$. Since we are looking for the probability that Smith is the first of the two to die, then we are evaluating $\Pr[X<Y]$. Let us plot the region of integration first, $0\le X\le5$, $0\le Y\le10$, and $X<Y$.

Based on the figure above, we can easily see that the corresponding region is bounded by $x<y<10$ and $0\le x\le5$, therefore \begin{align} \Pr[X<Y]&=\int_{x=0}^5\int_{y=x}^{10}f_{X,Y}(x,y)\ dx\ dy\\ &=\int_{x=0}^5\int_{y=x}^{10}f_{X}(x)\cdot f_{Y}(y)\ dx\ dy\quad\Rightarrow\quad\text{$X$ and $Y$ are independent}\\ &=\int_{x=0}^5\int_{y=x}^{10}\frac15\cdot \frac1{10}\ dx\ dy\\ &=\int_{x=0}^5\int_{y=x}^{10}\frac1{50}\ dx\ dy. \end{align} We obtain the result as the solution manual stated.

Answer 2

Let $X$ be Smiths' time of death and let $Y$ be Jone's time of death. You want to calculate $P(X<Y)$. To be able to calculate a probability concerning two different random variables you need their joint distribution/density. Luckily they are independent, so that $f_{(X,Y)}(x,y) = f_X(x)f_Y(y)$. The next step is to integrate this density over all possibilities $(x,y)$ where $x\le y$. This region is what is drawn in your book and in integral form it looks like your statement.

Answer 3

In probability we look for $P(X=x,Y>x)$ for all possible x. We fix $X=s$ and then integrate $Y$ from all values above $s$; we do this for each possible "$x=s$", hence "$ds$" from $(0,5)$.

You might think of it in discrete terms, where you would just fix each x and look at the corresponding y>x probability in each x-case with the x-probability and sum up. This summation is here put as integral for the continuous case.