# Finding a probability function R (the remainder when divided by a distribution)

• Oct 22nd 2011, 02:47 PM
KelvinScale
Finding a probability function R (the remainder when divided by a distribution)
Hi everyone, I am really unsure of how to do this question.

Let X have a geometric distribution with f(x) = p(1-p)^x for x = 0,1,2,....
Find the probability function of R, the remainder when X is divided by 4.

I don't know where to start, but I'm suspecting that you would construct a geometric series with r, and find an expression for f(r)... Not really sure.

• Oct 22nd 2011, 04:10 PM
emakarov
Re: Finding a probability function R (the remainder when divided by a distribution)
If $\displaystyle 0\le r < 4$, then $\displaystyle P(R = r) = P(X = r) + P(X = 4 + r) + P(X = 8 + r) + \dots =$$\displaystyle p(1 - p)^r + p(1 - p)^{4+r} + p(1 - p)^{8+r} + \dots$, which is a geometric series.
• Oct 22nd 2011, 04:16 PM
KelvinScale
Re: Finding a probability function R (the remainder when divided by a distribution)
So taking that geometric series, I can find the sum of the geometric series,which will express P (R = r)?

That is, we have P (R = r) = p(1-p)^r/ 1-((1-p)^4+r)) ?

THanks!
• Oct 22nd 2011, 04:31 PM
emakarov
Re: Finding a probability function R (the remainder when divided by a distribution)
Quote:

Originally Posted by KelvinScale
That is, we have P (R = r) = p(1-p)^r/ 1-((1-p)^4+r)) ?

There is no +r in the exponent: $\displaystyle P(R = r) =\frac{p(1-p)^r}{1-(1-p)^4}$.