1. ## Joint Probability

Is this correct?

1). X and Y are independent standard exponential random variables with $\displaystyle \lambda$ = 1
(a) Find $\displaystyle P(X\geq4, Y<3)$
I did $\displaystyle P(X\geq4, Y<3) = P(X\geq4)P(Y<3) = e^{-4}e^{-3} = e^{-7}$
(b) Find $\displaystyle E(X^2Y)$
(2) A random point (X,Y) is distributed uniformly on the square with vertices (1,1), (1,-1), (-1,1), and (-1,-1). The joint pdf is $\displaystyle f(x,y)=1/4$ on the square.
(a) Find $\displaystyle P(X^2 + Y^2 < 1)$
(b) Find $\displaystyle P(2X - Y > 0)$

If someone can just teach me how to do one of the above in (2), I think I can handle the other. Thank you!!

2. The pdf for an exponential distribution is exp (-x). To find the expectation we multiply by x and integrate to get 1/. To find the expectation of X^2 just square the pdf and multiply that by x and integrate. We get ( ^2)/4. Since X and Y are independent the expectation of the product is the product of the expectation. ie 1/4 (Note the subtlety X^2 IS dependent on X)

With uniform distribution this becomes a geometry problem. 2a is asking what portion of the square is covered by a circle of radius 1. 2b asks what portion of the square lies under the graph y=2x

3. Hello,
Originally Posted by VENI
Is this correct?

1). X and Y are independent standard exponential random variables with $\displaystyle \lambda$ = 1
(a) Find $\displaystyle P(X\geq4, Y<3)$
I did $\displaystyle P(X\geq4, Y<3) = P(X\geq4)P(Y<3) = e^{-4}e^{-3} = e^{-7}$
False.
$\displaystyle P(X<x)=1-e^{-x}$
So $\displaystyle P(X>x)=e^{-x}$

hence $\displaystyle P(X \geq 4)P(Y<3)=e^{-4} ({\color{red}1-}e^{-3})$

(b) Find $\displaystyle E(X^2Y)$
Since X and Y are independent, then X² and Y are independent.
Thus $\displaystyle E(X^2Y)=E(X^2)E(Y)$

(2) A random point (X,Y) is distributed uniformly on the square with vertices (1,1), (1,-1), (-1,1), and (-1,-1). The joint pdf is $\displaystyle f(x,y)=1/4$ on the square.
(a) Find $\displaystyle P(X^2 + Y^2 < 1)$
$\displaystyle x^2+y^2<1 \Rightarrow \{x \in (-1,1) ~,~ y \in (-\sqrt{1-x^2}~,~\sqrt{1-x^2})\}$
Hence $\displaystyle P(X^2+Y^2<1)=P(X \in (-1,1) ~,~ Y \in (-\sqrt{1-x^2}~,~\sqrt{1-x^2}))$

Now remember this definition (I'm sorry, I'm trying to explain so it may be a bit long) :
The cdf of a rv M is defined as : $\displaystyle F_M(m)=P(M\leq m)=\int_{-\infty}^m f_M(t) ~dt$, where $\displaystyle f_M$ is the pdf.
So $\displaystyle P(M \in (a,b))=\int_a^b f_M(t) ~dt$

So we have here :
$\displaystyle P(X^2+Y^2<1)=\int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} f(u,v) ~ dv ~du$, where f is the joint pdf

As bob000 said, you could also solve it by geometry. And I could explain the above thing more clearly by using indicator functions, but I don't know if you know it.