Product of random independent variables

Question

What are the properties of the product of random variables? The book I have on probability and statistics only comments on their sum properties.

I know that when you sum random independent variables you get to also sum their expected values (E) and their variance (Var). Is there anythinhg similar for product?

The problem I'm trying to solve is the following:

When you throw a dice, if the result if 1,2 or 3 you score 1, if the result is 4 you score 2, if it is 5 or 6 you score 3. Being N the number of dice throws necessary for the product of my scores being over 100000, what is P[N>25]?

(The problem comments that it is useful to use the normal distribution in this case)

I was trying to solve this by getting the expected value and variance of a dice throw, but I'm not sure if I can then later just multiply the variance and expected value. Am I going the right way? How should I proceed?

Answer 1

The hard way to answer this question is to compute all terms of the distribution of $N$ up to and including $P(N=25)$, add those probabilities and subtract the sum from $1$.

An easier way is to consider what would happen if you simply committed to rolling that same die $25$ times in succession, then compare the outcomes to the outcomes of the original problem.

Let the value of the $k$th roll be $X_k$ ($1 \leq X_k \leq 3,\ 1 \leq k \leq 25$) in your modified experiment. Any sequence of throws for which $N=17$ (for example) in the original problem will correspond to an event in your modified experiment in which $\prod_{i=1}^{16} X_i < 100000$ ($\prod_{i=1}^{16} X_i$ is the product of the first $16$ rolls) and $\prod_{i=1}^{17} X_i > 100000$. Moreover, the probability that $N=17$ in the original experiment is the same as the probability that $\prod_{i=1}^{16} X_i < 100000 < \prod_{i=1}^{17} X_i$ in the modified experiment.

But if $\prod_{i=1}^{17} X_i > 100000$, then $\prod_{i=1}^{25} X_i > 100000$ as well. In fact, any outcome of the modified experiment corresponding to a value of $N$ with $N \leq 25$ will imply that $\prod_{i=1}^{25} X_i > 100000$. On the other hand, the outcome $N \geq 26$ in the original experiment corresponds to the outcome $\prod_{i=1}^{25} X_i < 100000$ in the modified experiment, and has the same probability.

So we really just need to find out, if you roll this die $25$ times, what is the probability that the product of the $25$ values of $X_k$ will be greater than $100000$?

Then you can use some of the ideas in the other answers and comments. Define $Y_k = \log X_k.$ Then $$\begin{eqnarray} P(Y_k = 0) &=& \frac12.\\ P(Y_k = \log 2) &=& \frac16.\\ P(Y_k = \log 3) &=& \frac13. \end{eqnarray}$$ What you want to find then is $P(\sum_{i=1}^{25} Y_k > \log 100000).$ The reason the problem text suggests use of a normal variable is presumably because by the Law of Large Numbers, the sum of sufficiently many iid variables is approximately a normal variable.

Answer 2

This is not a complete answer to your question, but it might help solving it.

The minimum number of throws required in order to get over $100000$ is $11$:

• Either you get $3^{11}=177147$
• Or you get $3^{10}\cdot2=118098$

So you basically need to calculate $1-\sum\limits_{k=11}^{25}P(N=k)$.

You can calculate $P(N=11)=(\frac13)^{11}\cdot\binom{11}{11}+(\frac13)^{10}\cdot\binom{11}{10}\cdot(\frac16)^{1}\cdot\binom{1}{1}$.

Answer 3

Yes there is something similar to the summation, for the product. If all your random variables are independent, then you just apply the log function $$E[\log\prod_i X_i]=E[\sum_i \log(X_i)]=\sum_i E[\log X_i]$$