Find $\mathbb{P}(X+Y<1)$ given the joint density function of the vector $(X,Y)$ distributed uniformly over a region $R$.

Question

Let $X$ and $Y$ be random variables with the vector $(X, Y)$ uniformly distributed on the region $R = \{(x, y) : 0 < y < x < 1\}$. Find $\mathbb{P}(X+Y<1)$. I am pretty sure that I calculated the pdf correctly:

$f_{X,Y}(x,y)= 2$ if $(x,y) \in R$, $0$ elsewhere.

To get the probability, I know I need to double integrate $f(x,y)$. I am not sure of the bounds though. Is the correct integration $\int_0^1 \int_0^{1-x} 2\,dy\,dx$, or should the upper bound be $x$ so that it is $\int_0^1 \int_0^x 2 \, dy \, dx$. Or are neither of these correct?

Answer 1

$P(X+Y <1)= 2 \int_0^{1}\int_0^{\min \{x,1-x\}} \, dy\,dx$ since we need $y <1-x $ as well as $y <x$. Write this as $2 \int_0^{\frac 1 2}\int_0^{x} \,dy\,dx + 2 \int_{\frac 1 2}^1 \int_0^{1-x} \,dy\,dx$. I will let you finish.

Answer 2

While routinely studying the distribution of the random variable $X+Y$ leads us to an answer (as mentioned in an answer above), for this problem you don't need to go through all those evaluation of integrals.

There is another more visually appealing solution to this problem as well as several other problems of similar kind. Assuming that you have not come across such an technique till date I am going to introduce to you this new method via this solution, leaving the remaining (non-probabilistic) computations for you.

Notice that $R = \{ ( x,y ) : 0 < y < x < 1 \}$ is basically the region strictly bounded by the line $x = y,$ the $X$ axis and the points $x=0$ and $x=1.$ Notice that $R$ is precisely the triangle whose vertices are $(0,0), (1,0) , (1,1).$

Here by ''strictly bounded'' I mean that none of the above mentioned boundaries of $R$ can be a part of $R;$ however this wouldn't matter since the boundaries anyway have probability measure zero.

Now, notice that for two real numbers $a,b$ the condition $a+b \ge 1$ is same as saying $a < 1-b$ and this in turn is same as saying that the distance between the point $(a,b)$ and the origin is at least as big as the distance between the point $(a,b)$ and the line $x = 1-y.$

Putting all this together we see that $(X,Y)$ is uniformly distributed in the region strictly bounded by the line $x=y,$ the $X$ axis and the points $x= 0,$ and $x=1,$ and the probability $\mathbf{P} \{ X+Y \ge 1 \}$ is precisely the (conditional) probability that the point $(X,Y)$ lies in a location inside $R$ which is more closer to the line $x = 1-y$ than to the origin, thus $(X,Y)$ lies in the triangle formed by vertices $(1,0), (1/2 , 1/2) , (1,1).$

Now just compare the areas of the triangle $R,$ and the triangle formed by the vertices $(1,0) , (1/2 , 1/2) , (1,1),$ and their ratio would give us $\mathbf{P} \{ X+Y \ge 1 \}$, and thus subtracting it from $1$ (the total probability) would give us the value for $\mathbf{P} \{ X+Y < 1 \};$ thus for instance the same thing won't be true if $(X,Y)$ has the gaussian distribution or something else.

Footnote:the fact that the probability can be derived as the ratio of triangles relies heavily on the given condition that $(X,Y)$ has a uniform distribution; for a much better understanding of the importance of uniform distribution in this solution observe that from the very density (or definition) of uniform distribution it follows that if $\mathcal{X}$ is a bounded connected region in the plane with the boundary curve of $\mathcal{X}$ being smooth (although this is not strictly necessary) and having a finite non-zero area, and if $\mathcal{X}_1, \mathcal{X}_2$ are two connected disjoint (that is, $\mathcal{X}_1 \cap \mathcal{X}_2 = \phi$) subsets of $\mathcal{X}$ with $\mathcal{X} = \mathcal{X}_1 \cup \mathcal{X}_2$ and if the boundary curves of $\mathcal{X}_1$ and $\mathcal{X}_2$ are smooth, then for a random variable $Z$ distributed uniformly over $\mathcal{X}$ one has $\mathbf{P} \{ Z \in \mathcal{X}_i \} = \dfrac{ \text{ area of } \mathcal{X}_i }{ \text{ area of } \mathcal{X}}.$