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Imagine to have two urns (A and B) with equal number of balls (N) numbered from 1 to N. Each extraction consists of taking a ball from urn A and one from urn B, and then multiplying the respective numbers. Balls are discarded after and not reentered. Let's make N Extraction, and calculate the sum of the N resulting products.

I found the expected value for the sum being:

$$\frac{N(N+1)^2}{4}$$

which is a peace of the information that I'm looking for. Does anybody have a clue of what distribution is Sum going to follow. It's going to be close to a normal for big Ns, but not exactly since it's goint to have a well defined Min and a Max. What about the variance and standard deviation of this thing?

Dan
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Manuel Venè
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1 Answers1

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In this answer the route to finding variance, based on bilinearity of covariance and symmetry.

You can write the sum as $S=P_1+\cdots+P_N$ where the $P_i$ are the products.

Then: $$\mathsf{Var}(S)=\mathsf{Cov}(P_1+\cdots+P_N,P_1+\cdots+P_N)=N\mathsf{Var}(P_1)+N(N-1)\mathsf{Cov}(P_1,P_2)$$

Try to find $\mathsf{Var}(P_1)$ and $\mathsf{Cov}(P_1,P_2)$ yourself.

I have to go now, and later I will come back to have a second look.

edit (continuation)

Let $P_{i}=X_{i}Y_{i}$ where $X_{i}$ denotes the $i$-th number from urn A and $Y_{i}$ denotes the $i$-th number of urn B.

Then all $X_{i}$ and $Y_{i}$ for every $i\in\left\{ 1,\dots,N\right\} $ the random variables $X_{i}$ and $Y_{i}$ have uniform distribution over $\left\{ 1,\dots,N\right\} $.

Further the random vectors $\left(X_{1},\dots,X_{N}\right)$ and $\left(Y_{1},\dots,Y_{N}\right)$ are independent.

$$\mathbb{E}P_{1}=\mathbb{E}X_{1}Y_{1}=\mathbb{E}X_{1}\mathbb{E}Y_{1}=\left(\mathbb{E}X_{1}\right)^{2}$$

$$\mathbb{E}P_{1}^{2}=\mathbb{E}X_{1}^{2}Y_{1}^{2}=\mathbb{E}X_{1}^{2}\mathbb{E}Y_{1}^{2}=\left(\mathbb{E}X_{1}^{2}\right)^{2}$$

$$\mathsf{Var}P_{1}=\mathbb{E}P_{1}^{2}-\left(\mathbb{E}P_{1}\right)^{2}=\left(\mathbb{E}X_{1}^{2}\right)^{2}-\left(\mathbb{E}X_{1}\right)^{4}$$

$$\mathsf{Covar}\left(P_{1},P_{2}\right)=\mathbb{E}P_{1}P_{2}-\mathbb{E}P_{1}\mathbb{E}P_{2}=\mathbb{E}X_{1}Y_{1}X_{2}Y_{2}-\left(\mathbb{E}X_{1}\right)^{4}=$$$$\mathbb{E}X_{1}X_{2}\mathbb{E}Y_{1}Y_{2}-\left(\mathbb{E}X_{1}\right)^{4}=\left(\mathbb{E}X_{1}X_{2}\right)^{2}-\left(\mathbb{E}X_{1}\right)^{4}$$

$$\mathsf{Var}\left(S\right)=N\left[\left(\mathbb{E}X_{1}^{2}\right)^{2}-\left(\mathbb{E}X_{1}\right)^{4}\right]+N\left(N-1\right)\left[\left(\mathbb{E}X_{1}X_{2}\right)^{2}-\left(\mathbb{E}X_{1}\right)^{4}\right]=$$$$N\left(\mathbb{E}X_{1}^{2}\right)^{2}+N\left(N-1\right)\left(\mathbb{E}X_{1}X_{2}\right)^{2}-N^{2}\left(\mathbb{E}X_{1}\right)^{4}$$

What remains now is finding $\mathbb{E}X_{1}$, $\mathbb{E}X_{1}^{2}$ and $\mathbb{E}X_{1}X_{2}$. I leave that to you.

drhab
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  • There something about the product of two uniforms here:

    https://math.stackexchange.com/questions/659254/product-distribution-of-two-uniform-distribution-what-about-3-or-more

    I guess this would work for P1... but what about P2? This might simply be to hard

     – Manuel Venè May 30 '18 at 12:24
  • It is possible to find the distribution of $P_1$ by brute force and for every $i\in{1,\dots,N}$ the distribution of $P_i$ is the same as the distribution of $P_1$. Things become difficult however if you want the distribution of $S=P_1+\cdots+P_N$. This especially because the $P_i$ are not independent. It will definitely not lead to a closed form. – drhab May 30 '18 at 13:16
  • What about an index of dispersion around the mean? I don't actually need the distribution, but a value proportional to the expected dispersion would definitely be enough for my needs. Any idea? I could easily get a monte carlo variance, but that just doesn't sound anything close to a proper solution. – Manuel Venè May 30 '18 at 13:43
  • Sorry, but I am not familiar with stuff like "dispersion around the mean". – drhab May 30 '18 at 14:18
  • Sorry, that may be an englishization of an expression common in my mother language. Here we go https://en.wikipedia.org/wiki/Statistical_dispersion – Manuel Venè May 30 '18 at 14:25
  • That is not your fault. It is more that I am familiar with probability theory but not so much with statistics. I provided a way to find the variance but will not go on with distribution. So for me it ends here. Good luck. – drhab May 30 '18 at 14:30
  • I'm sorry, I didn't see the edit to your answer. Extremly deep and complete! – Manuel Venè May 30 '18 at 14:48
  • Glad to help. Check me on mistakes though. – drhab May 30 '18 at 14:50