1
$\begingroup$

We draw 2 numbers from a normal (gauss) distribution with mean $\mu$ and variance $\sigma$ and we add them to find the first value $a_1$ of a sequence. The second value $a_2$ of this sequence is the product of $a_1$ and 2 other numbers randomly drawn from the same probability distribution. The third value $a_3$ is the product of $a_2$ by 2 other randomly drawn numbers, and so forth....

Therefore: $a_1=X_1+X_2$ and $a_2=a_1\cdot (X_3+X_4)=(X_1+X_2)(X_3+X_4)$

What is the expected value of $a_n$?

Similar question here

UPDATE

Here is a plot of a very simple simulation where $a_n$ equals $a_{n-1}$ multiplied by a number drawn from a gaussian distribution with parameters $\mu=2$ and $\sigma=0$ for the red line and $\sigma=0.5$ for the blue line. I can run this simulation 100 times, in 95% of the cases, I get this kind of pattern. Therefore it seems that the variance is important to predict a value in a sequence. Is it correct?enter image description here blue = high variance | red = low variance | y-axis is logarithmic

| cite | improve this question | | | | |
$\endgroup$
  • $\begingroup$ To make sure I'm clear, let $\{X_i\}_{i=1}^{\infty}$ be a sequence of i.i.d. $N(\mu,\sigma^2)$ random variables. Then we can interpret $a_1 = X_1 + X_2$. Then $a_2 = a_1 X_3 X_4 = (X_1+X_2)X_3X_4$ or $a_2 = a_1(X_3+X_4) = (X_1+X_2)(X_3+X_4)$? $\endgroup$ – Tom Oct 30 '13 at 13:10
  • $\begingroup$ @Tom I think you're clear! The expected value of $a_1$ is the of the gaussian distribution. The expected value of $a_2$ is lower than $\mu^2$ if the $\sigma$ is higher than zero. Does it make sense? $\endgroup$ – Remi.b Oct 30 '13 at 13:21
  • $\begingroup$ @Remi.b You did not answer the question Tom asked. $\endgroup$ – Did Oct 30 '13 at 13:23
  • $\begingroup$ My confusion lies in whether $a_2=(X_1+X_2)X_3X_4$ or $a_2=(X_1+X_2)(X_3+X_4)$. That is, we pick two random numbers and add to make $a_1$. Then we pick two more random numbers, but do we then add these two $X_3+X_4$ before multiplying by $a_1$ resulting in $a_1(X_3+X_4)$? Or do we just multiply $a_1$ by $X_3$ and $X_4$ resulting in $a_2 = a_1X_3X_4$? $\endgroup$ – Tom Oct 30 '13 at 13:24
  • $\begingroup$ @Tom oh sorry. $a_2=a_1\cdot (X_3+X_4)=(X_1+X_2)(X_3+X_4)$ is correct. $a_2=a_1\cdot X_3\cdot X_4)$ is wrong $\endgroup$ – Remi.b Oct 30 '13 at 13:26
1
$\begingroup$

As already mentioned, $E[a_n]=(2\mu)^n$ for every $n$, whatever the variance is. An explanation for the deviation observed in the simulations might be the following (although I fail to see how the graph in the post could represent any sequence $(a_n)$ generated as the OP explains, since almost surely $a_n\lt0$ for infinitely many $n$... unless one plots $|a_n|$ and not $a_n$?).

Assume that $(x_n)$ is i.i.d. normal with mean $1$ and variance $v$, and consider $y_n=x_1x_2\cdots x_n$, then $y_n$ is $a_n/(2\mu)^n$ for some variance $v$ and indeed, $E[y_n]=1$. But the almost sure behaviour of $y_n$ (the one simulations would exhibit) is quite different.

To wit, considering the normal density $f_v$ with mean $1$ and variance $v$, one sees that $x_k$ is in the infinitesimal interval $(x,x+\mathrm dx)$ approximately $nf_v(x)\mathrm dx$ times hence $$ \sum_{k=1}^n\log|x_k|\approx\int\log|x|\,nf_v(x)\mathrm dx. $$ This heuristics can be made rigorous, which shows that, when $n\to\infty$, $|y_n|=\mathrm e^{nI(v)+o(n)}$, where $$ I(v)=\int_\mathbb R\log|x|\,f_v(x)\mathrm dx=\frac1{\sqrt{2\pi}}\int_\mathbb R\log|1+x\sqrt{v}|\,\mathrm e^{-x^2/2}\mathrm dx. $$ Thus, the ratio between the blue curve and the red curve should behave like $\mathrm e^{nI(v)}$. Qualitatively, this explains the straight lines in the simulations.

Quantitatively, the parameters used in the simulation (mean $2$, variance $1/2$) correspond to $v=1/8$, and $I(1/8)\approx-.0825$, which is negative, hence indeed the ratio goes to zero, almost surely. For $n=800$, this indicates a ratio red/blue of order $10^{29}$, which is much greater than what the simulations indicate.

My guess is that the parameters are actually mean $2$ and standard deviation $1/2$ (not variance). This guess yields $v=1/16$, $I(1/16)\approx-.0352$ and a ratio red/blue of order $10^{13}$, which seems compatible with the simulations.

| cite | improve this answer | | | | |
$\endgroup$
0
$\begingroup$

Let $\{X_i\}_{i=1}^{\infty}$ be i.i.d. $N(\mu,\sigma^2)$ random variables (representing the random numbers drawn from your distribution; here $\mu = mean$ and $\sigma^2=var$). Then we have $$ a_1 = X_1 + X_2,\, a_2 = (X_1 + X_2)(X_3+X_4), \,...,\, a_n = (X_1+X_2)(X_3+X_4) \cdots (X_{2n-1}+X_{2n}).$$ Now, using independence, $$ E[a_n] = \underbrace{E[X_1+X_2]E[X_3+X_4]\cdots E[X_{2n-1}+X_{2n}]}_{n \text{ factors}} = \big(E[X_1]+E[X_2]\big)^n=2^n\mu^{n}. $$ You should note that by assumption $ E[X_i]=\mu$ for each $i$, so $$E[X_i + X_{i+1}]=E[X_{i}]+E[X_{i+1}]=2\mu$$ which is where the $2\mu$ came from.

| cite | improve this answer | | | | |
$\endgroup$
  • $\begingroup$ If one multiply $x$ by $x$, he gets $x^2$. If he multiplies $(x+e)\cdot (x-e)$ he gets $x^2-e^2$. Therefore, I would think that the variance of the distribution from which the values are drawn is important. Am I wrong? Henry seemed to agree with this here $\endgroup$ – Remi.b Oct 30 '13 at 13:43
  • $\begingroup$ If the random variable $\{X_i\}_{i=1}^{\infty}$ are truly independent, then the above calculation is correct for expected value. The actual values that are realized by each of the $X_i$ are do depend on the variance; small variance implies that you're much more likely to draw near the average, a large variance implies that the drawn values can be quite a distance away from the average. If you're simulating something, how close your simulated values are to the average depends on the variance, but it's not that the average is wrong, its that the simulated values can be far away from the average. $\endgroup$ – Tom Oct 30 '13 at 13:56
  • $\begingroup$ I updated my question. You might want to have a look! Thank you for your help! $\endgroup$ – Remi.b Oct 30 '13 at 14:18
  • $\begingroup$ Let's ignore that we're looking at a sequence for a moment. Suppose that you wanted to simulate just a single random variable $X$ with average $E[X]=\mu$. If you have a large variance and you simulate it, then each value you get will likely vary quite a bit from the true value of $E[X]$. Does that mean that we need to calculate the variance to calculate $\mu$? No. But it means that it will take many more simulations for us to be confident that average of those simulations is close to $\mu$. See what I'm saying? Maybe I don't understand what you're asking..? $\endgroup$ – Tom Oct 30 '13 at 14:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.