# What is the probability that the product of $20$ random numbers between $1$ and $2$ is greater than $10000$?

Twenty random real numbers $a_1,a_2,\dots,a_{20}$ are chosen such that $1\le a_i \le 2$. What is the probability that their product is greater than $10000$?

(By random, I mean each real number in the interval $[1,2]$ has an equal chance of being chosen. All twenty numbers are chosen independent of each other.)

• I think you should ask for a general solution for the product of $N$ i.i.d. $a_i \sim U[1,2]$ such that the product of $a_i$ is less than some $1<K < 2^N$ – jameselmore Apr 23 '15 at 12:53
• That's equivalent to finding the probably that the sum of $N$ uncorrelated numbers $a_i \sim \log U[1,2]$ (in the range [0, log 2]) is less than $\log K$, which can be found by doing lots of convolutions. – kennytm Apr 23 '15 at 12:58
• "Each real number...has an equal chance of being chosen": You can't do that! What would this chance be? It can't be non-zero, and it seems like it can't be zero either (although technically it is). You have to say something like, "for any interval $[r,s] \subset [1,2$], the chance that $a_i \in [r,s]$ is $s-r$." This is called the uniform distribution on $[1,2]$. – TonyK Apr 23 '15 at 12:58
• @kennytm That's similar to what I thought, but I'm finding that the integrals need to be computed in a very piecewise manner, so not tractable for n=20. – cantorhead Apr 23 '15 at 13:03
• @John Smith, Does an approximation suffice via CLT, or do you mean an exact answer? – cantorhead Apr 23 '15 at 13:05

Here´s an aproximation with a normal distribution.

$${Let\quad Y_i=\ln( X_i) \quad where \quad X_i \thicksim U(1,2) \quad i=1,2, \ldots,20 } \qquad {then \quad f_{Y_i}(y_i)=e^{y_i}\quad 0\leq{y_i}\leq \ln(2)}$$

$$And \space then\quad {\mu =\int_0^{\ln(2)} ye^y dy =0.386294}\qquad{E[y^2]=\int_0^{\ln(2)} y^2 e^y dy =0.188317 }\qquad{\sigma=0.197722}$$ $$Now,\space from\space probability\space we\space know \space that:\qquad {\lim_{n\to \infty} Pr(Z_n= \dfrac{S_n-n\mu}{\sigma \sqrt n}\leq z )=\Phi (z)}$$ $$You´re\space looking\space for \qquad {\prod_{i=1}^{n} X_i \geq C }\qquad$$$${This\space means\qquad S_n=\sum_{i=1}^{n} \ln(X_i)=\ln(\prod_{i=1}^{n} X_i) \geq \ln(C) }\qquad$$$${Applying\space the \space C.L.T\space theorem:}$$$$\qquad \qquad {P(S_n \geq \ln(C))=P(S_n -n \mu \geq \ln(C) - n \mu)=P(\dfrac{S_n-n\mu}{\sigma \sqrt n}\geq \dfrac{\ln(C)-n\mu}{\sigma \sqrt n})={1-P(\dfrac{S_n-n\mu}{\sigma \sqrt n}\leq \dfrac{\ln(C)-n\mu}{\sigma \sqrt n})}\approx 1-\Phi( \dfrac{\ln(C)-n\mu}{\sigma \sqrt n})}$$ In your case n=20 (akward, since it´s not big enough to normal-approximate) and C= 10,000.

So the probability (approximately ) is 4.6596% (Montecarlo simulations says 4.51%).

Using the CLT on the logs as suggest by @kennytm might give a satisfactory approximation. (Of course the CLT does not apply directly to products.)

As long as we're discussing approximations, a simulation gave approximately 0.045 as the answer. Repeated it several times to check computational stability. One run is shown below. (Each row of the matrix has 20 observations from UNIF(1, 2); the penultimate statement takes products of rows; the last statement estimates the desired probability baed on a million performances of the experiment.)

 m = 10^6; n = 20
MAT = matrix(runif(m*n, 1, 2), nrow=m)
y = apply(MAT, 1, prod)
mean(y > 10000)
## 0.045291


Addendum: A histogram of the sums of logs of 20 obs is nicely fit by a normal density curve. This seems to lend credibility to the approach of using the CLT on logs.

• I think that an answer that uses a simulation should explain why you think the simulation is accurate, and if possible a confidence interval. – DanielV Apr 23 '15 at 17:13
• With a standard error of about $2$ in the ten-thousandths digit? – Brian Tung Apr 23 '15 at 17:13

This is more thinking aloud than real answer.

A paper called Product of n independent uniform random variables by Carl P. Dettmann, Orestis Georgiou seems to be relevant. Its central result is following theorem:

In our case $a=1$, $b=2$, $n=20$.

After a lot of cumbersome but doable calculations it is possible to obtain the probability from the question using PDF from above Theorem 1.