Suppose $X_i$ are independent random variables uniformly distributed on $[1,3]$. We are interested in the product $W=X_1X_2\cdots X_{10}$. Each $X_i$ is centered about $2$ so we might think $W$ should be centered about $2^{10}$.
Change the product into a sum with logs, so $Y_i=\ln(X_i)$. find the expected value of $Y_i$ and $\text{Var}(Y_i)$.
Use the Central Limit Theorem to estimate the prob $P(W<2^{10})$. What is the probability that the product will be as large as 2000? (Use above)?

I was able to figure out that the $E(Y_i)=\ln(3)/2$ and the $\text{Var}((\ln(3)^2)/12)$. However how do use those parts to solve $P(W<2^{10})$ and the probability that the product will be as large as $2000$?

  • 1
    $\begingroup$ We have $E(Y_i)=\int_1^3 \frac{\ln(x)}{2}\,dx$. An antiderivative of $\ln x$ is $x\ln x-x$, which does not seem to give your computed value. $\endgroup$ May 6 '15 at 22:59
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    $\begingroup$ I think that is the expected value of $X_i$. The bounds of $Y$ are from 0 to $ln(3)$. But I see how my above is incorrect. $\endgroup$ May 7 '15 at 20:13
  • $\begingroup$ No. The expected value of $X$ would be $\int_1^3 xf(x)\,dx = \int_1^3 x/2\,dx.$ $\endgroup$
    – BruceET
    May 9 '15 at 22:03

First, @Andre Nicolas is correct that you need to look again at your computations of the mean and variance of $Y$. Approximate values are $E(Y) \approx 0.648$ and $V(Y) \approx 0.0949.$ You should do the integration yourself to get slightly more accurate values.

Then $S = \sum_{i=1}^{10} Y_i$ has $E(S) \approx 6.48$ and $V(S) \approx 0.949.$

Now, you want to find
$$P(W < 2^{10}) = P(\ln W < 10\ln 2) = P(S < 6.9315).$$

By the CLT, $S$ is approximately $N(6.48, 0.949),$ so $P(S < 6.9315) \approx 0.6785.$ You might get a slightly better answer using more accurate values of $E(W)$ and $V(W).$ You can find $P(W > 1000)$ similarly.

The distribution of $Y$ is slightly skewed with no tails and so a sum of ten values should be nearly normal. Even so, a normal approximation is unlikely to give more than two places of accuracy, if that.

A simulation in R of 100,000 values of $W$ is given below. It shows that $W$ has mean near 1027, variance near 1144, and $P(W < 1024) \approx 0.671,$ which is likely more accurate than the normal approximation. The simulation error can be made arbitrarily small by increasing the number of simulated values.

A histogram (not included here) shows that the distribution of $W$ is heavily right-skewed with a mode near (but not at) 0 and very little probability beyond 5000.

 MAT = matrix(runif(10^6,1,3),ncol=10) # 100k x 10
 w = apply(MAT, 1, prod)  # multiplies across rows
 mean(w < 2^10)  # approximates P(W < 1024)
 ## 0.67104
 mean(w)         # approximates E(W)
 ## 1026.577
 sd(w)           # approximates SD(W)
 ## 1143.775
 mean(w < 5000)  # approximates P(W < 5000)
 ## 0.98575
 mean(w > 1000)
 ## 0.33765

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