Wikipedia:Reference desk/Archives/Mathematics/2013 February 11

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

February 11

Variance of the product of two normals

I know that for independent random variables, $Var(XY)=Var(X)Var(Y)+Var(X)[E(Y)]^{2}+Var(Y)[E(X)]^{2}$.

Does this mean that the variance of the product of $X\sim N(0,\sigma _{X}^{2})$ and $Y\sim N(0,\sigma _{Y}^{2})$ is simply $\sigma _{X}^{2}\sigma _{Y}^{2}$?

Thorstein90 (talk) 01:22, 11 February 2013 (UTC)

With all respect, a person who has the sophistication to ask that question should have the sophistication to answer it, since the product of two zero-mean Gaussians is a zero-mean Gaussian. Looie496 (talk) 04:36, 11 February 2013 (UTC)

Oh really? Then what is this: http://mathworld.wolfram.com/NormalProductDistribution.html Thorstein90 (talk) 04:47, 11 February 2013 (UTC)

At a glance that looks okay to me. In equation 2 of your wolfram reference, the expression $(|u|)/(\sigma _{x}\sigma _{y})$, looks to me like the ratio of the centered random variable to the standard deviation, so the variance would be $\sigma _{x}^{2}\sigma _{y}^{2}$. Duoduoduo (talk) 14:54, 11 February 2013 (UTC)

However, the product of two zero mean Gaussians is not Gaussian. That's true for sums not products. Duoduoduo (talk) 14:59, 11 February 2013 (UTC)

The answer to your (original) question is "yes". McKay (talk) 05:52, 11 February 2013 (UTC)

No: It's not Gaussian. Michael Hardy (talk) 02:07, 16 February 2013 (UTC)

The original question didn't say anything about Gaussians. -- Meni Rosenfeld (talk) 17:59, 21 February 2013 (UTC)

The number of subsets of {1,2,...,n} with a total sum k

What's the generation function $f_{n}$ of the series $a_{k}=\left|\left\{A\subseteq \{1,2,\dots ,n\}:\sum _{i\in A}i=k\right\}\right|$? Thanks, 79.179.188.147 (talk) 11:56, 11 February 2013 (UTC)

See Restricted partition generating functions in partition (number theory). Gandalf61 (talk) 13:19, 11 February 2013 (UTC)

Still can't solve it. The function $q(k)$ is similar but doesn't restrict the subsets $A$ to be contained in $\{1,2,\dots ,n\}$. 79.179.188.147 (talk) 14:17, 11 February 2013 (UTC)

The restriction to subsets of $\{1,2,\dots ,n\}$ means that the upper bound of the product is not infinity but is ... what ? Gandalf61 (talk) 14:31, 11 February 2013 (UTC)

Oh, thanks! 79.179.188.147 (talk) 17:11, 11 February 2013 (UTC)

The number of ways to color N balls with I colors is. ${\binom {N+I-1}{I-1}}$. Bo Jacoby (talk) 02:51, 12 February 2013 (UTC).

Wouldn't the answer be (1 + x)(1 + x²)⋅⋅⋅(1 + xⁿ)? If you expand the product, the coefficient of x^k is a_k. 64.140.122.50 (talk) 04:56, 13 February 2013 (UTC)

Squared triangular number representative polychora.

Not sure if this belongs here or on WP:MATH. In the article Squared triangular number, is the following text:

These numbers can be viewed as figurate numbers, a four-dimensional hyperpyramidal generalization of the triangular numbers and square pyramidal numbers.

. Now this makes it sound like there is a single 4 dimensional figure that if sliced differently into two dimensional pieces would make it hypergeometically obvious. While the sum of cubes is obviously a cubical hyperpyramid (shape made of a cube and 6 square pyramids), it is unclear to me how the hyper pyramid would be cut to give the square of the triangular number. Can someone please help with explaining this to me here, which might help in changing the explanation in the article?Naraht (talk) 15:08, 11 February 2013 (UTC)

Don't know of any, but there is a nice proof without words at [1] Dmcq (talk) 01:04, 13 February 2013 (UTC)

Not doubting the proof (though some more information as to which proof used by each of the discoverers might be nice. Part of the problem with the concept is that the cubical hyperpyramid has a single vertex which is unique so figuring out which of the 6 it lines up with.Naraht (talk) 19:01, 13 February 2013 (UTC)

Have you tried examining the cross-sections involving planes involving the w-axis? They are hard to see. Double sharp (talk) 15:50, 18 February 2013 (UTC)

There are two types of cross section on the cubical hyperpyramid. The ones from the unique vertex to the largest cube are essentially from the point to the largest cube with a cube getting larger. In the other three cross sections, it starts with a square, builds up to a square pyramid and then back down to a square. Not sure how that helps. :(Naraht (talk) 15:43, 20 February 2013 (UTC)