# Calculate $P(X \leq 1 \cap Y \leq 1)$ and $f_{Y|X}(y|x=0)$ of a density function [closed]

I have a density function with

$f(x, y) = x^3+0.25y$, $0 \leq x \leq 1, 0 \leq y \leq 2$

and for x and y out of that range it is defined as 0.

First i need to calculate $P(X \leq 1 \cap Y \leq 1)$. How can i do that? The variables X and Y are not indepent. I calculated

$f_X(x) = 2x^3 + 0.5$ and $f_Y(y) = 0.25+0.25y$.

What is my next step?

Now i have to calculate

$f_{Y|X}(y|x=0)$ , with $f_X(x) = 2x^3+0.5$, $0 \leq x \leq 1$.

What do i have to do here? I really don´t have any idea.

(This are just exam preparation excercises)

## closed as off-topic by Did, Namaste, Shailesh, Leucippus, José Carlos SantosJul 17 '17 at 1:27

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• isn't it just a double integral: integrate over the rectangle? – user29418 Jul 15 '17 at 21:11

Hints

1: $P(X\leq 1, Y\leq 1)=\int_0^1 \left( \int_0^1 x^3+0.25y \ \, dy \right) \, dx$

2: $f_{X|Y}(y|x)=\frac{f(x,y)}{f_X(x)}$

Then set $x=0$ to obtain $f_{Y|X}(y|x=0)$

• Thank you very much sir. – Nu Ta Jul 15 '17 at 14:51
• You are welcome. Feel free to accept the answer. – callculus Jul 15 '17 at 14:53

For the first part note that $\{ x\le 1 \cap y \le 1\} = D = \{(x,y)| x\in [0,1], y \in [0,1]\} = [0,1]^2$, thus $$P((x,y)\in D) = \int \int_{D} f_{XY}(x,y)dxdy.$$ As $D$ is a simple rectangle, the order of integration does not matter.

For the second part, just use the definition of conditional probability, $$P(A|B) = \frac{P(A\cap B)}{P(B)} \to f_{Y|X}(y|x) = \frac{f_{X,Y}(x,y)}{f_X(x)}.$$