1. Multivariate Probability Distribution

Question: f(x, y) = 1, where x,y are between 0 and 1 OR = 0, elsewhere
a) What is P(X - Y > 0.5)?
b) What is P(XY < 0.5)?

Honestly I have no idea how to go about this question. Can anybody point out how to get it started?

2. Re: Multivariate Probability Distribution

Originally Posted by Greenmango
Question: f(x, y) = 1, where x,y are between 0 and 1 OR = 0, elsewhere
a) What is P(X - Y > 0.5)?
b) What is P(XY < 0.5)?

Honestly I have no idea how to go about this question. Can anybody point out how to get it started?
First, define $X,Y$. Once you know that, you can start working on the problem.

3. Re: Multivariate Probability Distribution

What do you mean by "define X and Y"? Find their marginal probability distributions?

4. Re: Multivariate Probability Distribution

Originally Posted by Greenmango
What do you mean by "define X and Y"? Find their marginal probability distributions?
Ah, I did not understand Multivariate Probability Distribution to be Marginal Probability Distribution. In general, $X,Y$ are random variables, not specifically related to $f$. Perhaps your book defines them canonically somehow to relate to $f$. Perhaps $X = \int_{-\infty}^x f(x,y)dy$ and $Y = \int_{-\infty}^y f(x,y)dx$?

Or is $f$ the probability distribution, letting you know that $0\le x,y \le 1$ with probability 1 and outside of that range with probability 0?

5. Re: Multivariate Probability Distribution

The second option, f(x,y) is the joint probability distribution for X and Y.

6. Re: Multivariate Probability Distribution

$$P(X-Y>0.5) = \int_{-\infty}^\infty \int_{-\infty}^{x-0.5}f(x,y)dydx = \dfrac{1}{8}$$

See what you can do for part B.

7. Re: Multivariate Probability Distribution

I am not really sure how to give a hint, so here is my solution:

If $x<0.5$ or $y<0.5$, then $XY<0.5$. Also if $x>0.5$ and $y>0.5$, you can still have $xy<0.5$ if $y<\dfrac{1}{2x}$.

Thus, you have:

\begin{align*}P(XY<0.5) & = P(X<0.5\text{ or }Y<0.5)+P( X>0.5\text{ and }Y>0.5\text{ and } XY < 0.5 ) \\ & = \dfrac{3}{4}+\int_{0.5}^1\int_{0.5}^{1/2x}dydx \\ & = \dfrac{3}{4}+\dfrac{1}{2}\left(\ln 2 - \dfrac{1}{2}\right)\end{align*}

So, you get

$$P(XY<0.5) = \dfrac{3+\ln\left(\dfrac{4}{e}\right)}{4}$$

8. Re: Multivariate Probability Distribution

Originally Posted by SlipEternal
$$P(X-Y>0.5) = \int_{-\infty}^\infty \int_{-\infty}^{x-0.5}f(x,y)dydx = \dfrac{1}{8}$$

See what you can do for part B.

\begin{align*}P(X-Y>0.5) &= \int_{-\infty}^\infty \int_{-\infty}^{x-0.5}1dydx \\
&= \int_{0}^1 \int_{0}^{x-0.5}1dydx \\
&= \int_{0}^1 (x - \frac{1}{2}-0)dx \\
&= (\dfrac{1}{2}x^2 - \dfrac{1}{2}x)|^1_0 \\
&= \dfrac{1}{2} - \dfrac{1}{2} \\
&= 0 \end{align*}

What did I mess up now?

9. Re: Multivariate Probability Distribution

Originally Posted by Greenmango
\begin{align*}P(X-Y>0.5) &= \int_{-\infty}^\infty \int_{-\infty}^{x-0.5}1dydx \\
&= \int_{0}^1 \int_{0}^{x-0.5}1dydx \\
&= \int_{0}^1 (x - \frac{1}{2}-0)dx \\
&= (\dfrac{1}{2}x^2 - \dfrac{1}{2}x)|^1_0 \\
&= \dfrac{1}{2} - \dfrac{1}{2} \\
&= 0 \end{align*}

What did I mess up now?
The lower bound for x needs to change. If $y<x-0.5$, then $0.5+y<x$, which means $0.5<x<1$.

10. Re: Multivariate Probability Distribution

Ah okay gotcha, thanks so much!