Uniform Bivariate Probabilities

Question

Suppose I have a uniformly distributed variable X on the interval $[0,a]$ and a uniformly distributed variable Y on the interval $[0,b]$. Thus we have the respected pdfs as $f(x)=\frac{1}{a}$ and $f(y)=\frac{1}{b}$. I want to figure out the probability that $P(X+Y = c)$. I tried interpetign it diagramatically, as for when one is dealing with the $\leq$ or $\geq$ problems (area of the function in the bounded region), but then I can not compare areas as $x+y=c$ has no area and thus I can't use the fact that $p=\frac{\text{area of the event occuring}}{\text{total area}}$. Moreover, I thought that since it doesn't have an area this would imply that the probability is $0$, however, there are infinitely many combinations each with a probability of $\frac{1}{a} \times \frac{1}{b}$ which suggests this is also false. Does anybody have an idea how to approach this?

Answer 1

As per the fact that X and Y are continuous rv's it is self evident that

$$\mathbb{P}[X+Y=c]=0$$

$\forall c$

It would be different if you wanted to derive the density of $X+Y$ but also in this case it is not possible to solve the problem unless you have enough information about the dependence structure between X and Y (i.e. X,Y independent)

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So Let's assume X,Y independent and let's suppose we are interested in calculating $f_Z(z)$, say the law of $Z=X+Y$ and let's suppose $a>b$

First let's note that $f_Z(z)$ is the pdf of the rv, say a probability density and not a "probability function". Known the pdf we can calculate the probability of any interval of Z.

To calculate f it is easier to calculate $F_Z(z)$ first, using the definition.

Before begin the calculation it is very useful to do a drawing of the problem

As you should know, the CDF of the sum is the area under the $Y=z-X$ line, where $z \in [0;a+b]$ multiplied by the joint density $f(x,y)=\frac{1}{ab}$

It is very easy to calculate it in the following way

$$ F_Z(z)=\mathbb{P}[Z \leq z] = \begin{cases} 0, & \text{if $z <0$} \\ \frac{z^2}{2ab}, & \text{if $0 \leq z <b$} \\ \frac{2z-b}{2a}, & \text{if $b \leq z <a$} \\ 1-\frac{(a+b-z)^2}{2ab}, & \text{if $a \leq z <a+b$} \\ 1, & \text{if $z \geq a+b$} \end{cases}$$

If you want to derive $f(z)$ you have only to derivate F. If you set $a=b$ the calculations are easier; if you set $a<b$ the brainstorming is the same but reversed.

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How F is calculated in the various intervals?

Just as an example, in the interval $b\leq z<a$ the CDF is

$$F_Z(b)+\frac{z-b}{ab}b=\frac{b}{2a}+\frac{z-b}{ab}b$$

that is (area of the first triangle + area of the parallelogram) on the total area $ab$

.... and so on

Answer 2

It is straightforward to use calculus to find the distribution of $X + Y.$ [The joint distribution of independent $(X,Y)$ is uniform on a rectangle.] Maybe this example, simulated in R, will help you visualize the problem

set.seed(2020)
x = runif(10^5, 0, 3)
y = runif(10^5, 0, 7)
hist(x + y, prob=T, col="skyblue2")

To find the CDF of $T = X + Y$ you will need to find, for example, $P(T \le 6).$

plot(x,y, pch=".")
 cond = (x + y <= 6)
 points(x[cond],y[cond], pch=".", col="blue")