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Sum of random number of random variables 
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#1
Jul3108, 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 


#2
Aug108, 05:54 AM

P: 284

Use the tower property which says that [tex] E(E(XY))=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.



#3
Aug108, 03:34 PM

Sci Advisor
P: 6,056

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.



#4
Aug108, 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~~ 


#5
Aug308, 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(XY)] + Var(E[XY])[/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_iN] = NE[X][/tex] [tex]Var(\sum_{i = 1}^{N}X_iN) = NVar(X)[/tex] But if I substitute these into the conditional variance formula I get: [tex]Var(X) = E[Var(XN)] + Var(E[XN])[/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? 


#6
Aug308, 02:03 PM

P: 532




#7
Aug308, 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). 


#8
Aug308, 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! 


#9
Feb2510, 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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