Hey guys, in this video I'm going to do this probability and statistics problem.
This is basically a continuation from the previous problem. Where in the
previous problem we were finding the marginal PMF from a joint
PMF, in this case in this problem we're going to find the marginal PDF from the
joint PDF. The difference is that this is a continuous distribution or these
are continuous variables and in the previous case when you have PMFs you're
dealing with discrete variables. We're going to do similar things, find a
probability in C and then find the marginals in D and E and this problem
has the additional part of A and B where we also are going to find the support
and the constant C for this function. Let's go ahead and get started. A says
sketch the support of X and Y. This just tells you where you're going to
have your limits. So basically where does this probability density function exist
in 2D space. This is going to become relevant for C and then D and E. In A
it might seem like it's a simple problem to just sketch but we're going to use
the sketch in later problems. So we have that and we're done. This is X and
this is Y and this is basically going to help us for the support. B says find C so
that f of X, Y is a valid joint PDF. For questions like this, the integral
over the whole support of the joint probability density function has to
equal 1. In this case we're just going to plug in that for 0 1 dx dy and 0 3, we
have C X plus C Y and we're going to solve that equal to 1 and solve for C.
I went ahead and distributed the C to the parentheses inside. Let's
integrate X first. That's going to be C X squared over 2 plus C Y X from 0 to 1
dy and then this is still dy and we're going to plug in 1. That's
going to give C half plus CY and then we still have to do the outer integral.
When we do that, we're going to get CY over 2 plus CY squared over 2. Integrate
from 0 to 3. And we get 3 halves C plus 9 halves C equal to 1.
That's going to give us 6 C is equal to 1, which is going to give us that C
is equal to 1/6 and we're done with B. Now C says find the probability that X
is less than Y. We're going to go back to the support and now we don't want the
complete--we don't want to integrate over the complete support because that
would give us 1, as we just found in B, but we want to find only the integral
that will give us when X is less than Y. When we do that integral it will give us
the probability. We're going to draw the line X is equal to Y which is just
from 0 to 1 and then this point is (1, 1). And because this is when X is less than
Y, the support is everything above this line. This whole shaded area is equal
to when X is less than Y. Now that's going to give us a picture of what we
need to integrate. It's going to be another double integral and there are
multiple ways--again this is just calc 2 I think--there are multiple ways of
getting this integral. You can either do 1 minus the triangle and that will
give you your answer or you can just do the integral of this area that's shaded
in. And again you can kind of switch in-- it doesn't matter the order if you
integrate dx or dy first. I did 0 to 1 X to 3 of 1/6 X plus 1/6 Y dy dx.
I'm integrating Y first and then dx and again that doesn't matter. And notice
that we're plugging in the value of C that we got for B. This double
integral, I'm not going to write it out but you're going to get 11/12. And if you
had done one minus the unshaded triangle, you would have gotten
1 minus 1/12 and gotten the same answer. So we're done with C. D and E say find the
marginal probability density function and the first one we're going to find is
the probability density function f of X. The formula for the continuous case
to find the marginal is that f of X is equal to the joint over dy. So every
instance of dy. In this case, this is going to be a single integral. That's
going to be from 0 to 3 which is again the support of Y and we're just going to
plug in the joint formula again for the PDF. That was 1/6 X plus 1/6 Y dy.
That is going to give us 1/6 XY plus Y squared over 12 from 0 to 3. When we plug
that in we're going to get 1/2 X plus 9/12 and this is over the existence of X.
So X is from 0 to 1. We do have to include that as part of our answer and
we're done with this prob--or sorry not this problem, we're done with D and
for E we're doing a similar formula. So f of Y, the marginal of Y from the joint XY,
is all the support when you integrate dx. So f of X,Y dx. In this case the support
we already know is from 0 to 1 for all of X and then we have 1/6 X plus 1/6 Y dx.
This gives us X squared over 12 plus XY over 6 from 0 to 1 and when we plug in 1
we're going to get 1/12 plus Y over 6 and this is over the whole support or
the whole existence of Y which is from 0 to 3. And now that's the final answer for
E and we are done with this problem.
*chiptune music*