Infinite coin flipping Show $ sum_{k=0}^{infty} f(k,r,p) = 1 $
For a infinite coin flip consider the probability of success $ p in (0,1) $.
For $n in mathbb{N}_0 $ and $ r in mathbb{N} $ define the probability $f(k,r,p)$ ( $k$ failures then the $r$th sucess ).
Show:
$ sum_{k=0}^{infty} f(k,r,p) = 1 $.
So I've already have a possible solution but I want to know if my solution is ok. So can you please check my attempt?
I want to ask a tiny question before I try to prove the equation: $f$ is a generalization of a geometric distribution, right? Anyways:
So $ sum_{k=0}^{infty} f(k,r,p) = sum_{k=0}^{infty} {k+r-1 choose k}p^r(1-p)^k = p^r sum_{k=0}^{infty} {k+r-1 choose k}(-1)^k(-(1-p))^k =p^r sum_{k=0}^{infty}{-r choose k}(-(1-p))^k = p^r(1-(1-p))^{-r} = p^rp^{-r} = 1.$
What did I do in the third equation? : ${k+r-1 choose k}(-1)^k = {-r choose k} $ .
calculus sequences-and-series probability-theory probability-distributions
asked Nov 21 at 21:26
RukiaKuchiki
299211
add a comment |
For a infinite coin flip consider the probability of success $ p in (0,1) $.
For $n in mathbb{N}_0 $ and $ r in mathbb{N} $ define the probability $f(k,r,p)$ ( $k$ failures then the $r$th sucess ).
Show:
$ sum_{k=0}^{infty} f(k,r,p) = 1 $.
So I've already have a possible solution but I want to know if my solution is ok. So can you please check my attempt?
I want to ask a tiny question before I try to prove the equation: $f$ is a generalization of a geometric distribution, right? Anyways:
So $ sum_{k=0}^{infty} f(k,r,p) = sum_{k=0}^{infty} {k+r-1 choose k}p^r(1-p)^k = p^r sum_{k=0}^{infty} {k+r-1 choose k}(-1)^k(-(1-p))^k =p^r sum_{k=0}^{infty}{-r choose k}(-(1-p))^k = p^r(1-(1-p))^{-r} = p^rp^{-r} = 1.$
What did I do in the third equation? : ${k+r-1 choose k}(-1)^k = {-r choose k} $ .
calculus sequences-and-series probability-theory probability-distributions
asked Nov 21 at 21:26
RukiaKuchiki
299211
1
Your argument is correct. About your tiny question, the distribution is known as Negative Binomial
– Daniel
Nov 21 at 22:14
add a comment |
For a infinite coin flip consider the probability of success $ p in (0,1) $.
For $n in mathbb{N}_0 $ and $ r in mathbb{N} $ define the probability $f(k,r,p)$ ( $k$ failures then the $r$th sucess ).
Show:
$ sum_{k=0}^{infty} f(k,r,p) = 1 $.
So I've already have a possible solution but I want to know if my solution is ok. So can you please check my attempt?
I want to ask a tiny question before I try to prove the equation: $f$ is a generalization of a geometric distribution, right? Anyways:
So $ sum_{k=0}^{infty} f(k,r,p) = sum_{k=0}^{infty} {k+r-1 choose k}p^r(1-p)^k = p^r sum_{k=0}^{infty} {k+r-1 choose k}(-1)^k(-(1-p))^k =p^r sum_{k=0}^{infty}{-r choose k}(-(1-p))^k = p^r(1-(1-p))^{-r} = p^rp^{-r} = 1.$
What did I do in the third equation? : ${k+r-1 choose k}(-1)^k = {-r choose k} $ .
calculus sequences-and-series probability-theory probability-distributions
asked Nov 21 at 21:26
RukiaKuchiki
299211
For a infinite coin flip consider the probability of success $ p in (0,1) $.
For $n in mathbb{N}_0 $ and $ r in mathbb{N} $ define the probability $f(k,r,p)$ ( $k$ failures then the $r$th sucess ).
Show:
$ sum_{k=0}^{infty} f(k,r,p) = 1 $.
So I've already have a possible solution but I want to know if my solution is ok. So can you please check my attempt?
I want to ask a tiny question before I try to prove the equation: $f$ is a generalization of a geometric distribution, right? Anyways:
So $ sum_{k=0}^{infty} f(k,r,p) = sum_{k=0}^{infty} {k+r-1 choose k}p^r(1-p)^k = p^r sum_{k=0}^{infty} {k+r-1 choose k}(-1)^k(-(1-p))^k =p^r sum_{k=0}^{infty}{-r choose k}(-(1-p))^k = p^r(1-(1-p))^{-r} = p^rp^{-r} = 1.$
What did I do in the third equation? : ${k+r-1 choose k}(-1)^k = {-r choose k} $ .
calculus sequences-and-series probability-theory probability-distributions
calculus sequences-and-series probability-theory probability-distributions
asked Nov 21 at 21:26
RukiaKuchiki
299211
asked Nov 21 at 21:26
RukiaKuchiki
299211
edited Nov 21 at 22:06
asked Nov 21 at 21:26
RukiaKuchiki
299211
asked Nov 21 at 21:26
RukiaKuchiki
299211
asked Nov 21 at 21:26
RukiaKuchiki
299211
299211
1
Your argument is correct. About your tiny question, the distribution is known as Negative Binomial
– Daniel
Nov 21 at 22:14
add a comment |
1
Your argument is correct. About your tiny question, the distribution is known as Negative Binomial
– Daniel
Nov 21 at 22:14
1
1
Your argument is correct. About your tiny question, the distribution is known as Negative Binomial
– Daniel
Nov 21 at 22:14
Your argument is correct. About your tiny question, the distribution is known as Negative Binomial
– Daniel
Nov 21 at 22:14
add a comment |
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Your argument is correct. About your tiny question, the distribution is known as Negative Binomial
– Daniel
Nov 21 at 22:14