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Sum of random number of random variables

by fredliu
Tags: number, random, variables
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fredliu
#1
Jul31-08, 03:35 PM
P: 2
Hi, Guys,
I'm new to this forum, and don't have strong background in probability theory, so please bare with me if the question is too naive.

Here's the question,

In a problem I'm trying to model, I have a random variable (say, R), which is a sum of random number (say, N) of random variables (say, Hi), in which all Hi are i.i.d..

I have distribution of both N and Hi, and I am interested in the expected value and variance of R.

Any suggestions how I can get it? My initial thought is E(R) = E(N)*E(Hi), but i feel it not quite right.. and it's even harder to have variance of R.

I did some googling, and found out ways to sum rvs, but not so much of how to find random sums..

Any suggestions? or hint about where I can find related information?

Thanks
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Focus
#2
Aug1-08, 05:54 AM
P: 284
Use the tower property which says that [tex] E(E(X|Y))=E(X) [/tex]. In your case the solution is [tex] E\left(\sum_{i=1}^N H_i \right)=E\left(E\left(\sum_{i=1}^N H_i | N\right)\right)=E\left(\sum_{i=1}^N E(H_i)\right)=E(N E(H_1)) [/tex]. Furthermore is N and His are independent then you can say that [tex] E(R)=E(N)E(H_1) [/tex]. Hope this helps.
mathman
#3
Aug1-08, 03:34 PM
Sci Advisor
P: 6,037
To get the variance, you can apply the same approach (as Focus) to get the second moment and then use the usual relationship between second moment and variance.

fredliu
#4
Aug1-08, 03:47 PM
P: 2
Sum of random number of random variables

Thanks very much for all your replies, guys~~

I'll look into the suggested approach, thanks a bunch~~
pluviosilla
#5
Aug3-08, 10:05 AM
P: 13
This question is related to another, so if I may, I'd like to add it to this thread.

In my Sheldon Ross, First Course in Probability, there is a derivation that has stumped me. The author wants to show how to use the conditional variance formula

[tex]Var(X) = E[Var(X|Y)] + Var(E[X|Y])[/tex]

to derive the following identity:

[tex]Var(\sum_{i = 1}^{N}X_i) = E[N]Var(X) + (E[X])^2Var(N)[/tex]

but he skips some steps and succeeds in losing me. :-) All he says, by way of derivation, is that the following two statements hold:

[tex]E[\sum_{i = 1}^{N}X_i|N] = NE[X][/tex]
[tex]Var(\sum_{i = 1}^{N}X_i|N) = NVar(X)[/tex]

But if I substitute these into the conditional variance formula I get:

[tex]Var(X) = E[Var(X|N)] + Var(E[X|N])[/tex]
[tex] = E[NVar(X)] + Var(NE[X])[/tex]
[tex] = E[N]E[Var(X)] + Var(NE[X])[/tex]

In the last step, I can separate E[N] because N and X are independent, but I can think of no further simplifications. I've been looking around for a handy identity for a variance of a product, but cannot find anything.

Suggestions?
gel
#6
Aug3-08, 02:03 PM
gel's Avatar
P: 532
Quote Quote by pluviosilla View Post
[tex]Var(X) = E[Var(X|N)] + Var(E[X|N])[/tex]
[tex] = E[NVar(X)] + Var(NE[X])[/tex]
[tex] = E[N]E[Var(X)] + Var(NE[X])[/tex]

...

Suggestions?
how about using [itex]E[Var(X)]=Var(X)[/itex] and [itex]Var(NE[X])=Var(N)E[X]^2[/itex]?
n_bourbaki
#7
Aug3-08, 02:12 PM
P: 103
E[X] is just a number, so you have to work out the variance of a constant times N. That is standard: Var(kY)=k^2.Var(Y) for k constant and Y an r.v.

Don't forget that Var (Y)= E(Y^2) - E(Y)^2 as well, when you're doing things like this. So if U and V are independent

Var(UV)= E(U^2V^2) - E(UV)^2 = E(U^2)E(V^2) - E(U)^2E(V^2)

which can be related, albeit messily, to Var(U) and Var(V).
pluviosilla
#8
Aug3-08, 05:55 PM
P: 13
"E[X] is just a number" - Yes! Seems obvious now, but that is what I was overlooking.

I suppose we could also say that Var(X) is just a number, which explains the other identity that I overlooked: E[Var(X)] = Var(X).

Thanks very much for these helpful replies!
subodha
#9
Feb25-10, 07:15 PM
P: 1
Please consider this one as well.

I have a set of (say N) random variables (X_i), of which I know the pmf. I want to find the probability of (sum(i=1 to N) X_i)=K where K is a consnant.


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