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.
$2.5\;$ Generate numerically $M$ random numbers $x$ each being the product of $N$ independent random numbers equally distributed between $1$ and $2$ with $M$ between $10^3$ and $10^6$ and $N$ between $5$ and $50$. Approximate the distribution of $x$ by a histogram and compare the evolution with $N$ of the most probable value $x_\text{mp}$ of $x$ with its average $\langle\!\langle x \rangle\!\rangle$ and the quantity $x_\text{typ} := \exp(\langle\!\langle \ln x \rangle\!\rangle)$. Could you give an argument why asymptotically (for $N \to \infty$) the most probable value and $x_\text{typ}$ should coincide? Why is the most probable value always smaller than $x_\text{typ}$ in the simulations? (Hint: Compare the equations determining the most probable value of $x$ and of $\ln x$.)
[First attempt to solve the question - by using the result provided here]
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:
$$\mu :=\text{mean}(\ln(Y))=\ln(4)-1=0,38629$$
and
$$V:=\text{variance}(\ln(Y))= 1-2\ln^2(2)=0.03909$$
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:
$$x_{mp}= e^{N\mu-NV}=e^{N(0.34)}\rightarrow \infty$$
$$x_{typ}= e^{N\mu}= e^{N(0.38)}\rightarrow \infty$$
