ter3k4w8x

2022-10-05

What is distribution of sum of two geometric distribution?

We are tossing an unfair coin (the probability of throwing a head is p). We count the number of the first identic throws (for example, if we have HHHT..., then this number is three). What is the distribution of this random number?

We are tossing an unfair coin (the probability of throwing a head is p). We count the number of the first identic throws (for example, if we have HHHT..., then this number is three). What is the distribution of this random number?

Alannah Hanson

Beginner2022-10-06Added 11 answers

Step 1

Note that it doesn't make sense for n to be 0. So the formula you found only works for $n\ge 1$.

Step 2

To show that the sum of the probabilities will be 1, we find the value of the sum

$\sum _{n=1}^{\mathrm{\infty}}{p}^{n}(1-p)+(1-p{)}^{n}p$

$=\sum _{n=1}^{\mathrm{\infty}}({p}^{n}(1-p)+(1-p{)}^{n}p)$

$=p(1-p)(\sum _{n=1}^{\mathrm{\infty}}{p}^{n-1}+\sum _{n=1}^{\mathrm{\infty}}(1-p{)}^{n-1})$

$=p(1-p)(\sum _{n=0}^{\mathrm{\infty}}{p}^{n}+\sum _{n=0}^{\mathrm{\infty}}(1-p{)}^{n})$

$=p(1-p)(\frac{1}{1-p}+\frac{1}{p})$

$=p+(1-p)$

$=\overline{){\displaystyle 1}}$

So we have shown that the sum of the probabilities will be 1. I don't believe there is a simpler expression for the formula you found.

Note that it doesn't make sense for n to be 0. So the formula you found only works for $n\ge 1$.

Step 2

To show that the sum of the probabilities will be 1, we find the value of the sum

$\sum _{n=1}^{\mathrm{\infty}}{p}^{n}(1-p)+(1-p{)}^{n}p$

$=\sum _{n=1}^{\mathrm{\infty}}({p}^{n}(1-p)+(1-p{)}^{n}p)$

$=p(1-p)(\sum _{n=1}^{\mathrm{\infty}}{p}^{n-1}+\sum _{n=1}^{\mathrm{\infty}}(1-p{)}^{n-1})$

$=p(1-p)(\sum _{n=0}^{\mathrm{\infty}}{p}^{n}+\sum _{n=0}^{\mathrm{\infty}}(1-p{)}^{n})$

$=p(1-p)(\frac{1}{1-p}+\frac{1}{p})$

$=p+(1-p)$

$=\overline{){\displaystyle 1}}$

So we have shown that the sum of the probabilities will be 1. I don't believe there is a simpler expression for the formula you found.

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