# find expectation of this r.v.

• Mar 17th 2013, 12:57 PM
pv3633
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
• Mar 17th 2013, 02:32 PM
ILikeSerena
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:

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

Wolfram|Alpha could not solve this (within its timeout).
But I found that a close numerical approximation is:

$EX \approx \frac 5 4 \sqrt n + \frac 3 4$
• Mar 17th 2013, 05:14 PM
pv3633
Re: find expectation of this r.v.
thank you.
that's the sum i got also.

how to get the approximation btw?
• Mar 17th 2013, 05:19 PM
ILikeSerena
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
• Mar 18th 2013, 04:51 PM
pv3633
Re: find expectation of this r.v.
thank you