Mean of a product of random variables
Comment: Although you do not need to find the distribution of the product, you might get a better intuitive grasp of the problem by looking at a related simulation. To keep the products from getting unmanageably large, I simulated 100,000 random samples of size $n = 5$ from $\mathsf{Unif}(6, 12).$ Thus $E(X_i) = 9$ and $E(P) = 9^5 = 59049.$ With 100,000 iterations it is reasonable to expect about two significant digits of accuracy.
m = 10^5; n = 5; x = runif(m*n, 6, 12)
MAT=matrix(x,nrow=m) # each row of matrix has sample of size 5
p = apply(MAT, 1, prod) # products of values in each row
mean(p)
## 59029.94 # aprx E(P) = 59049
Here is a histogram of the simulated distribution of $P$ with a vertical red line at the mean.
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user12321
Updated on August 01, 2022Comments
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user12321 over 1 year
I am having difficulty with a question that asks me to find the mean of P, where P = $X_1X_2X_3X_4X_5$. $X_1,X_2,X_3,X_4,X_5$ are a random sample from a uniform distribution on [600,1200]. Any help would be greatly appreciated.
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StubbornAtom over 5 yearsDon't forget to tell us what you have tried and where exactly are you stuck.
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user12321 over 5 yearsSorry I am new to this. So far I have looked at getting the expectation of P, this then brings me to a situation where I find myself having to integrate the product of 5 random variables and that is how far I have got. But even at that, I am not sure if I am thinking correctly about this. correctly.
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StubbornAtom over 5 yearsIt is given that it is a random sample. That makes it easy. Do you know what random sample implies?
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user12321 over 5 yearsYes I know what it means. I was unsure if it implies independence
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StubbornAtom over 5 yearsBy definition, $X_i$'s are independent and identically distributed random variables, $i=1,2,\cdots,5$. No reason to be unsure.
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Remy over 5 yearsNo integration is needed. Hint: $E(XY)=E(X)E(Y)$ if $X$ and $Y$ are independent. Do you know what the expected value of a uniform random variable is with parameters $a$ and $b$?
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user12321 over 5 yearsYes its .5(a+b), so I just have (.5(a+b)) to the 5th power?
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Remy over 5 yearsYes, that is correct.
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user12321 over 5 yearsI'm sorry, this is probably a really silly question but is the mean not 3?
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BruceET over 5 yearsMean of $X \sim \mathsf{Unif}(600,1200)$ is $E(X) = 900.$ Mean of $ \mathsf{Unif}(6,12)$ is 9. In your Question $E(P) = 900^5,$ as in one of the Comments. Which random variable has mean 3? Maybe I'm not getting your question.
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user12321 over 5 yearsI mistakenly thought the formula was .5(a-b) instead of .5(a+b). I really appreciate this explanation, thank you.