find expectation of this r.v.

toss balls one at a time into n bins, each ball will always land in one of the n bins.

stop tossing once some bin end up with 2 balls. and the tosses are independent of each other

X be the number of tosses needed.

(so X is between 2 and n+1)

find E(X)

i find linearity hard to apply here.

and the naive definition resulted in a messy sum, that i cannot reduce to simple form.

is this a well know distribution somewhere?

thanks for any insights

Re: find expectation of this r.v.

Quote:

Originally Posted by

**pv3633** toss balls one at a time into n bins, each ball will always land in one of the n bins.

stop tossing once some bin end up with 2 balls. and the tosses are independent of each other

X be the number of tosses needed.

(so X is between 2 and n+1)

find E(X)

i find linearity hard to apply here.

and the naive definition resulted in a messy sum, that i cannot reduce to simple form.

is this a well know distribution somewhere?

thanks for any insights

Hi pv3633! :)

This is not a well known distribution as far as I know.

The formula for EX is:

$\displaystyle EX=\sum_{k=2}^{n+1} k \cdot \frac {n!(k-1)!} {(n-k+1)!n^k}$

Wolfram|Alpha could not solve this (within its timeout).

But I found that a close numerical approximation is:

$\displaystyle EX \approx \frac 5 4 \sqrt n + \frac 3 4$

Re: find expectation of this r.v.

thank you.

that's the sum i got also.

how to get the approximation btw?

Re: find expectation of this r.v.

Quote:

Originally Posted by

**pv3633** thank you.

that's the sum i got also.

how to get the approximation btw?

I calculated a couple of values with Wolfram|Alpha:

Code:

`n EX`

1 2

10 4.6

100 13.2

1000 40.3

10000 125.66

Then I made a log-log-plot in excel which showed a straight line.

I used excel's solver to find the coefficients.

The resulting approximation is:

Code:

`n EX Approx`

1 2 2

10 4.6 4.702847075

100 13.2 13.25

1000 40.3 40.27847075

10000 125.66 125.75

Re: find expectation of this r.v.