Most probable value vs typical value in a specific simulation

could you please give me some hint on how to tackle the following question? I think I am not on the right way as I am thinking that deriving the pdf of a product of $N$ uniformly distributed random variables on the interval [1,2] is mandatory for the solution. The question is from this book.

I know that the most probable value is the one that makes the derivative of the pdf equal to zero. But I do not get what the author exactly means with his hint.

Thank you!

[EDIT - First attempt to solve the question]

Ok, so I got that the key of the question is to make use of the Central Limit Theorem.

We have $X=Y_1 \dotsb Y_N$ where $Y$ are iid and are uniformly distributed on $[1,2]$.

Hence, by the central limit theorem, $\ln(X)=\ln(Y_1)+\dotsb+\ln(Y_N)$ is approximately normally distributed for $N$ sufficiently large.

So, lets calculate the mean and vaiance of $\ln(Y)$. As $Y\sim U[1,2]$, it is immediate that $\ln(Y)\sim e^y \,,y\in [0,\ln(2)]$. Hence, after some integration we have the first two moments of $\ln(Y)$. They are:

$$\mu :=\text{mean}(\ln(Y))=\ln(4)-1$$

and

$$V:=\text{variance}(\ln(Y))= \ln^2(2)-\ln^2(4)+3$$

Hence, $\ln(X)\sim N(N\mu,NV)$.

As $\ln(X)$ is normally distributed, then X is lognormally distributed. More preciselly $X\sim \text{logN}(N\mu,NV)$.

Hence, we have everything that we need to finish the exercise.

The author asks for a comparasion among $x_{mp}$ and $x_{typ}$. So lets do it:

$x_{mp}:= \text{mode}(X)= e^{N\mu-NV}$ That is the mode that wikipedia gives for the lognormal distribution.

$$x_{typ}:=e^{\mathbb{E}(\ln(X))}= e^{N\mu}$$.

Finally, as $N \rightarrow \infty$, we have:

$$x_{mp}= e^{N\mu-NV}=e^{N(-1.17)}\rightarrow 0$$

$$x_{typ}= e^{N\mu}= e^{N(0.38)}\rightarrow \infty$$

The conclusion is that the author asks to verify that when $N\rightarrow\infty$, $x_{mp}$ and $x_{typ}$ coincide. But what I am finding is exactly the contrary. I think I am missing some detail. Could you please give me some more help?

[EDIT - Second attempt to solve the question]

$$\frac{1}{N}\ln(X)=\frac{1}{N}\left[\ln(Y_1)+\dotsb+\ln(Y_N)\right]$$ Hence: $$\frac{1}{N}\ln(X) \sim N(\mu,V/N)$$ Hence: $$\ln(X)\sim N(N\mu,NV)$$ And I am on the same way as before... I did not get what it changes to divide the sum by N. Could you please give me some more help, I think I am very close to the solution but I still cannot get it.

Edit to expand on the answer: On the log scale since the distribution approaches the normal distribution the mode which you refer to as the most probable value will approach the mean. Let Y represent the sum of the N variables on the log scale. Let X = exp(Y). Then X is the product random variable on the original scale. The monotonicity of the exponential function should help you get inequalities about the mean and mode on the original scale. Also ln(X) / N = ln(X$^1$$^/$$^N$).
• Thank you! Very helpful insight with the log, wich makes the variables $X$ sum, and consequently makes it easy to get the pdf (and the mean and variance). But for the product of the $X$ I see how to get the expected value but I do not see how to get the variance... Is there a way to imply a range for the most probable value (mode) from the mean and variance? – AnUser May 11 '17 at 14:24