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#1




Conditional expectation problem
I am having trouble figuring this problem. f(x,y)=1/(10x) 0<=y<=x<=10 What is E[xy=3]=?
For the numerator I substituted y=3 in the range of x and integrated x/(10x) from x=3 to x=10. Is the numerator done? Or is there another step? For the denominator I substititued y=3 in the range of x and integrated 1/(10x) from x=3 to x=10 to get f(y). Is the denominator done? Or is there another step? 
#2




It sounds like what you did was correct. What was your answer?
You have a joint distribution defined on the triangular region with vertices . The probability that for any region in that triangle is the volume bounded by the curve , the xyplane and this region projected to all z. Knowing , your probability that x falls in any range would naturally be reduced to the area under the curve with y fixed at 3... but there's a problem because that area isn't 1, in general, over all valid x! So to create a valid PDF that is proportional to 1/(10x) in this region from 3 to 10 (since those are the values permitted with y=3), you must divide by the right constant to make it integrate to 1. That's the denominator you're describing (the marginal distribution of Y evaluated at 3). Once you divide by that constant and you now have a valid PDF, you can just find the expected value with the usual definition. Hopefully that description helps this make more sense for you. Also the fact that might make this all feel a bit weird... how can you condition on an event with probability 0, right? In actuality, in the continuous case here you're conditioning on a 2D proximity of y=3 and taking the limit as that region shrinks to y=3... but I guess you don't need to know all that. Last edited by Z3ta; 11192017 at 05:19 PM.. 
#3




I never got an answer. I got this question on a trial exam. I didn't figure an answer at the time. And somehow skipped over the solution to this question. So, I couldn't check how it is supposed to be done. It took me awhile to figure out to do it the way I described, but had no way to check my solution. I was hung up on the fact that it was y=3 as opposed to y<3 or y>3, because as you said P(y=3) is technically 0. Also, in the fact that y was not part of f(x,y) unless you take the fact that y is part of the domain of x. You normally have to do a double integral for the numerator and denominator. But in this case I substituted y=3 in the domain of x and only did a single integral for both the numerator and denominator, so that was confusing me also. The answer I get is .7/(ln10ln3). Sound right?

#4




The 1/10 cancel in the numerator and denominator so I think the answer was
Your answer is about 0.58 which wouldn't make sense since you're finding the expected value of a random variable that falls between 3 and 10 always. You also know that 1/(10x) decreases with x so the mean necessarily has to be below the midpoint of 3 and 10 (6.5). In other words, the density is concentrated to the left so the mean would be too. Yea it's confusing when you learn this stuff in the continuous case since you're conditioning on a zero probability event. Most study resources never resolve this. I remember I was very confused when I first learned it for the exam. In the continuous case you're really using old conditional notation as a shorthand for a limit. You don't have to consider that to actually solve it, since all of this is hidden within integral properties and the fact that integrals are defined with limits. However, I find that if I don't know what's under the hood sotospeak it can make me second guess myself and negatively impact my confidence when doing problems. EDIT: Also the fact that the joint PDF is only a function of x just means the height of your surface (joint PDF) only depends on x. All of the same methods work just as well. You could even have a surface that is the same height everywhere and depends neither on x or y. Last edited by Z3ta; 11192017 at 06:47 PM.. 
#5




I can see where you're right about the 1/10 in the denominator. I left that out. Oops. The fact that you only do a single integral was hard for me to get my head around. Because normally you're supposed to integrate x over all values of x And integrate y over all values of y. I'm thinking of it like this: Since y can only have a single value, substituting y=3 in the range of x before integrating x is simultaneously integrating over all the values of y. Thanks for the help. You really helped clear this problem (and hopefully future ones) up for me.

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