For the second the density of is the convolution of the densities of and , which gives the triangular distribution on (symmetric about )
RonL
Hello everyone
Would appreciate help with these two, especially the second one.
The joint densities of and is given by:
where and are independent exponential random variables with parameter
Prove that S and T are independent by finding and and showing that
Once again I'd also like to ask if someone can suggest a book with sufficient examples and explanations regarding this kind of work. The book I have has very few examples, and even just as few explanations.Let and be independent and uniform on . Find and sketch the density function of
Still a little bit lost on this one. The lecturer said we'll hear about convolutions a bit later, and not to worry about it yet... (I'll have to go look up what a triangular distribution is as well.)
This has helped a lot, thank you Captain.
Here's just a guess... they're exponential so they must be positive?
EDIT: Wait, for some bizarre reason I also think they must be both negative. I tried working it out by myself just now...
To Jan:
A reference you might find helpful:
http://www.dartmouth.edu/~chance/tea...k/Chapter7.pdf
Given your current knowledge, I think it’s instructive for you to think about how to calculate the pdf of .
The cdf of T is given by since Y > 0.
Therefore:
where is the joint pdf of X and Y.
Since X and Y are independent random variables: for x > 0, y > 0, and is zero elsewhere.
Therefore:
.
Therefore the pdf of T is given by .
This is (of course) in agreement with the marginal distribution found using the given joint pdf of S and T:
.....
After studying the definition of convolution and looking at CaptainB’s example you should attempt to get the pdf of S = X + Y.
To Jan: It can also be done without ever knowing that you're essentially doing a convolution. To simplify the notation, I'll use S = U + V.
Since U and V are independent, their joint pdf is g(u, v) = (1)(1) = 1 for , , and zero elsewhere.
The cdf of S is given by .
There are two cases to consider: and .
Case 1: .
.
Therefore for .
Case 2: .
I'll leave the mopping up to you. You should get for .
And of course f(s) = 0 elsewhere. These are the same results as CaptainB got using the convolution theorem.