00:01
In this question, we have two random variable x and y, and we are given the conditional density function for y at x.
00:22
Let's say it is equal to x times e to the power negative xy, y is greater than zero and it is equal to zero else.
00:38
Okay, under this condition, we want to find the probability of y less than two, conditioning on x being two.
00:54
Okay, then by our definition, it is actually equal to the integral or y less than, the region determined at y less than two, and the integral over the conditional function for y conditioning on x being two, dy.
01:19
Okay, so the first thing for us is to plug two into our expression to find this density.
01:31
And by our previous expression, we know this density is equal to, now as x is equal to two times e to the power, by negative two y.
01:46
Okay, and by our definition, y need to be greater than zero, so the integral goes from zero to two, two times e to the power negative two y, dy, which is equal to e to the power negative two y, and y evaluated at two and zero, one minus e to the power negative four.
02:17
Okay, this is the first part.
02:21
The second part, we are required to compute the conditional expectation for y under this guy.
02:29
The same thing, this is the integral for the whole space.
02:35
I mean, y goes from zero to positive infinity.
02:37
Y times this density, okay, which is equal to two times e to the power negative two y, dy.
02:52
And now combine those long term, we know it is equal to d negative e to the power minus two y.
03:04
And use the integration by part, times minus y evaluated at positive infinity and zero, minus negative e to the power dy.
03:27
Okay, when y goes to positive infinity, this term goes away, because this term is much smaller than this guy.
03:36
So the whole term goes away.
03:38
Y goes to zero, then this term goes away.
03:42
So we get, doing some conservation, zero to positive infinity, e to the power negative two y, dy.
03:54
Add a two here and divide it back.
04:00
Now this term is equal to times negative e to the power negative two y, y evaluated at positive infinity and zero.
04:26
So it is equal to one over two.
04:27
The whole thing is equal to one over two.
04:37
Let me just do something to give us more space.
04:55
For the fourth part, we are required to compute the conditional expectation for any number x.
05:13
Okay, by our assumption, as x is a uniform distribution on zero or ten.
05:22
So we know here, we need to assume x to be greater than zero and less than ten.
05:31
Okay, and this expectation is only meaningful in this range.
05:36
Okay, by the definition, this is equal to, integral goes from zero to positive infinity, y times the conditional density, right? okay, that is equal to y times x, times e to the power negative xy, dy.
06:04
Okay, do the same thing as we just want to integrate everything with respect to y.
06:09
So we can regard x as a constant.
06:12
Combining these terms again, it is equal to d negative e to the power negative xy.
06:21
Okay, then using integration by power, y times negative e to the power negative xy.
06:30
Well, y is positive infinity and zero and plus, because we can do some consolation with the minus sign, just as what we've done previously.
06:41
So it can be written as e to the power negative xy, dy.
06:50
Okay, notice x is greater than zero by our assumption.
06:55
So when y goes to positive infinity, this term just goes away.
07:00
When y goes to zero, then this term is equal to zero...