# Thread: Bivariate Probabilty - Two r.v.s uniformly distributed over a triangle...

1. ## Bivariate Probabilty - Two r.v.s uniformly distributed over a triangle...

Hi all,

I'm trying to do the following problem and I'm having trouble figuring out how to set up my integrals:

Suppose that $Y_1$ and $Y_2$ are uniformly distributed over the triangle defined by the following points on y1,y2:
(-1,0), (0,1), (1,0)

a) Find $P(Y_1\leq{3/4}, Y_2\leq{3/4})$

What I've been able to do up to the point I'm confused:
Because it's uniformly distributed over a triangle, I found that the probability is uniformly distributed over area: bh/2 = 2*1/2 = 1
Therefore, I get the distribution function:
$f(y_1, y_2) = \begin{cases}1&\text{ if -1}\leq{y_1}\leq{1}, 0\leq{y_2}\leq1 \\0 &\text{otherwise}\end{cases}$

Now I can't figure out how to write the double-integral, can I just use one or do I have to find the sum of different components of the part of the triangle that fits the constraints?

Hi all,

I'm trying to do the following problem and I'm having trouble figuring out how to set up my integrals:

Suppose that $Y_1$ and $Y_2$ are uniformly distributed over the triangle defined by the following points on y1,y2:
(-1,0), (0,1), (1,0)

a) Find $P(Y_1\leq{3/4}, Y_2\leq{3/4})$

What I've been able to do up to the point I'm confused:
Because it's uniformly distributed over a triangle, I found that the probability is uniformly distributed over area: bh/2 = 2*1/2 = 1
Therefore, I get the distribution function:
$f(y_1, y_2) = \begin{cases}1&\text{ if -1}\leq{y_1}\leq{1}, 0\leq{y_2}\leq1 \\0 &\text{otherwise}\end{cases}$

Now I can't figure out how to write the double-integral, can I just use one or do I have to find the sum of different components of the part of the triangle that fits the constraints?
I suggest that you draw on a set of $Y_1 Y_2$-axes the region defined by the triangle.

Then shade that part of the region corresponding to $Y_1 < \frac{3}{4}$ and $Y_2 < \frac{3}{4}$. Since the joint pdf is uniform, the area of this region will give the probability and you don't need to set up and solve a double integral.

However, if you do want to integrate, note that the region is defined by the lines $Y_2 = 0$, $Y_2 = Y_1 + 1$ and $Y_2 = -Y_1 + 1$ (taking $Y_2$ as the vertical axis).

3. Thanks Mr. Fantastic,

I understood everything you said and have already shaded the triangle before asking the question (I just couldn't display it here).

Is it necessary to break this into 3 seperate parts over which I can integrate and add together?

I tried doing this:
$
P(Y_1\leq{3/4}, Y_2\leq{3/4}) = \int^{-1/4}_{-1} \int^{1+y_1}_{0}dy_2dy_1 +\int^{1/4}_{-1/4}\int^{3/4}_{0}dy_2dy_1 + \int^{3/4}_{1/4} \int^{1-y_1}_{0}dy_2dy_1
$

I got the correct answer 29/32.

Q1) Am I making it more complicated than it has to be by dividing it into 3?
Q2) If I want to find $P(Y_1-Y_2\geq{0})$
I just do:
$\int^{1}_{1/2} \int^{1-y_1}_{0}dy_2dy_1$
which gives 1/8 but the answer key in my book says 1/4, this must be a type right?

Thanks Mr. Fantastic,

I understood everything you said and have already shaded the triangle before asking the question (I just couldn't display it here).

Is it necessary to break this into 3 seperate parts over which I can integrate and add together?

I tried doing this:
$
P(Y_1\leq{3/4}, Y_2\leq{3/4}) = \int^{-1/4}_{-1} \int^{1+y_1}_{0}dy_2dy_1 +\int^{1/4}_{-1/4}\int^{3/4}_{0}dy_2dy_1 + \int^{3/4}_{1/4} \int^{1-y_1}_{0}dy_2dy_1
$

I got the correct answer 29/32.

Q1) Am I making it more complicated than it has to be by dividing it into 3?
Q2) If I want to find $P(Y_1-Y_2\geq{0})$
I just do:
$\int^{1}_{1/2} \int^{1-y_1}_{0}dy_2dy_1$
which gives 1/8 but the answer key in my book says 1/4, this must be a type right?
Is this a probability problem or a calculus problem?

Just do as Mr Fantastic says, draw a diagram, shade the area's you want the probability for. As the area of the original triangle is 1, you just want the area's of the regions defined. These are all easilly determined using the triangle area formula.

This is a demonstration of understanding not of your ability to set up and evaluate multi-variable integrals over odd-shaped regions.

See the attachment, the area in question for Q2 is shaded. The area of the main triangle is 1, the shaded area is 1/4 of the big triangle so its area ia 1/4.

CB

Thanks Mr. Fantastic,

I understood everything you said and have already shaded the triangle before asking the question (I just couldn't display it here).

Is it necessary to break this into 3 seperate parts over which I can integrate and add together?

I tried doing this:
$
P(Y_1\leq{3/4}, Y_2\leq{3/4}) = \int^{-1/4}_{-1} \int^{1+y_1}_{0}dy_2dy_1 +\int^{1/4}_{-1/4}\int^{3/4}_{0}dy_2dy_1 + \int^{3/4}_{1/4} \int^{1-y_1}_{0}dy_2dy_1
$

I got the correct answer 29/32.

Q1) Am I making it more complicated than it has to be by dividing it into 3?
Q2) If I want to find $P(Y_1-Y_2\geq{0})$
I just do:
$\int^{1}_{1/2} \int^{1-y_1}_{0}dy_2dy_1$
which gives 1/8 but the answer key in my book says 1/4, this must be a type right?
What CaptainBlack has said is quite right and is what you should do.

Originally Posted by CaptainBlack
[snip]This is a demonstration of understanding not of your ability to set up and evaluate multi-variable integrals over odd-shaped regions.[snip]
is particularly pertinent.

However, I suspect that you won't be happy until you know why you got the wrong answer when you attempted to set up and solving a double integral.

If you integrate first with respect to $y_2$ then the integral has to be done over two seperate regions. Go back and look at the diagram and you'll see this.

That being the case, it's easier to integrate first with respect to $y_1$:

$\Pr(Y_1 - Y_2 \geq 0) = \Pr(Y_2 \leq Y_1) = \int_{y_2 = 0}^{1/2} \int_{y_1 = y_2}^{y_1 = 1 - y_2} 1 \, dy_1 \, dy_2 = \frac{1}{4}$

as it should. Note that the intersection point of $y_2 = y_1$ and $y_2 = -y_1 + 1$ is $\left( \frac{1}{2}, \, \frac{1}{2} \right)$.