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 independently of each other.)

Solution 1:

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.

Solution 2:

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%).