Urn problem (indisting. objects into distinguishable urns)

[fignewtons]

Homework Statement

I have n balls and m urns numbered 1 to m. Each ball is placed randomly and independently into one of the urns.
Let X_i be the number of balls in urn number i.
So X₁+...+X_m = n
What is the distribution of each X_i?
What is EX_i and VarX_i
What is E[X_iX_j] given i≠j
What is Cov(X₁,X_j?

Homework Equations

Cov(XY)=E_xy(XY)-E_x(X)E_y(Y)

The Attempt at a Solution

I read: https://www.artofproblemsolving.com/wiki/index.php?title=Distinguishability and identified this as the last case.
I understand that there (n+m-1)ℂ(m-1) ways to place the balls but not how to describe this in the form of a pdf so that I can find expectation and variance and such.

 

[Ray Vickson]

Homework Statement

I have n balls and m urns numbered 1 to m. Each ball is placed randomly and independently into one of the urns.
Let X_i be the number of balls in urn number i.
So X₁+...+X_m = n
What is the distribution of each X_i?
What is EX_i and VarX_i
What is E[X_iX_j] given i≠j
What is Cov(X₁,X_j?

Homework Equations

Cov(XY)=E_xy(XY)-E_x(X)E_y(Y)

The Attempt at a Solution

I read: https://www.artofproblemsolving.com/wiki/index.php?title=Distinguishability and identified this as the last case.
I understand that there (n+m-1)ℂ(m-1) ways to place the balls but not how to describe this in the form of a pdf so that I can find expectation and variance and such.

Clearly, for a single urn, the distribution of the number $X_i$ in urn i is the same for any i = 1,2, ...,n, so we might as well look at urn 1. For urns i and j ≠ i the bivariate distribution of $(X_i,X_j)$ is the same for any pair i and j, so we might as well look at urns 1 and 2.

To find the marginal distribution of $X_1$, just look at it as a problem having two urns: 1 and not-1. For each object, the probability it goes into urn 1 is 1/m, while the probability it goes into urn not-1 is (m-1)/m.

For urns 1 and 2 look at it as a three-urn problem with urns 1, 2 and not-12. For each object, p(1) = p(2) = 1/m and p(not-12) = (m-2)/m.