Suppose and are two independent variables that follow a uniform distribution in [0,1]. calculate $ ( + > 2)$

Question

Suppose $$ and $$ are two independent variables that follow a uniform distribution in [0,1], calculate:

a. $ ( + > 2)$

b. $(+>5√)$

Well, i don't get what is this question tries to say. does it mean i should find the pair of $X$, $Y$ which for example meet $ + > 2$ condition? in this case the probability of part a would be zero right? how can i solve it in formulas?

And is there a way i can show it visually? like matlab?

Answer 1

I'll go through an example for the first one, but it will depend on your level of insight into probability theory.

First you need the joint law of $X$ and $Y$. The density of a uniform random variable on $[0,1]$ is given by $f(x) = \mathbf{1}_{[0,1]}(x)$. Since the two variables are independant, the joint density is just the product of both densities. Then $P(X+Y > 2) = \int_0^1 \int_0^1 \mathbf{1}_{x + y > 2} dx dy = 0$.

Of course it is the complicated way to prove something obvious, as mentioned by lulu in the comment. However this method can be generalised, for instance to compute the second one. This also gives an interpretation geometrically: you are looking at the area of the square $[0,1] \times [0,1]$ that satisfies the requirement in your probability, which might be the quickest way to compute the second one (and the easiest numerically).

Answer 2

The joint density function is $f(x,y)=1$ for $(x,y)\in [0,1]^2$ and zero elsewhere.

For part $A$ you have to integrate over the intersection of the square and above the triangular region bounded by $x+y=2$ which is empty so you get zero as you have indicated. Graph is helpful to show the regions.