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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.

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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.

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It might help to look at the distribution of the log first. For each variable the log will have independent and identically distributed distribution on [log(1), log(2)] = [0, log(2)] Note that the distribution of the log is not uniform. But since you are taking the log these variables sum. Derive the distribution for the log and then the normalized sum will approach a normal distribution. So you can get the limiting distribution. Note that for each N you can compute the mean and variance for the sum. I think this will help you.

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$).

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  • $\begingroup$ 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? $\endgroup$ – AnUser May 11 '17 at 14:24
  • $\begingroup$ @AnUser Please take note of my additions to my answer which I think partially address the questions in your comment. $\endgroup$ – Michael R. Chernick May 11 '17 at 15:14
  • $\begingroup$ I tried to solve the problem in the edit, but I need some more help... Could you please ckeck my edit? $\endgroup$ – AnUser May 11 '17 at 17:42
  • $\begingroup$ You need to normalize which involves dividing the sum by N in determining the mean and the most probable value. $\endgroup$ – Michael R. Chernick May 11 '17 at 17:48
  • $\begingroup$ Could you please point in which step I should divide by N? I think i did not get your last comment. My mistake is on the application of the Central Limit Theorem? $\endgroup$ – AnUser May 11 '17 at 17:54

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