1. ## Naming the distribution

I am given the scenario where a biased coin, with probability p € (0,1) of landing on heads is tossed repeatedly. For each n=1,2,..... let Xn be the total number of heads seen after the nth toss.
What is the distribution of Xn?

Is Xn a geometric distribution?

Thanks

2. ## Re: Naming the distribution

Originally Posted by yellowcarrotz
I am given the scenario where a biased coin, with probability p € (0,1) of landing on heads is tossed repeatedly. For each n=1,2,..... let Xn be the total number of heads seen after the nth toss. What is the distribution of Xn?
I hope that I am reading you correctly.
$\left( {\forall n \geqslant 1} \right)\left[ {\;X_n = k,\,k = 0,1, \cdots ,n} \right]$.
In other words, $X_n$ can have $n+1$ values.

The probability distribution is binomal: $\mathcal{P}\left( {X_n = k} \right) = \binom{n}{k}p^k(1-p)^{n-k}$

Did I understand what you are asking?

3. ## Re: Naming the distribution

Originally Posted by Plato
I hope that I am reading you correctly.
$\left( {\forall n \geqslant 1} \right)\left[ {\;X_n = k,\,k = 0,1, \cdots ,n} \right]$.
In other words, $X_n$ can have $n+1$ values.

The probability distribution is binomal: $\mathcal{P}\left( {X_n = k} \right) = \binom{n}{k}p^k(1-p)^{n-k}$

Did I understand what you are asking?

The geometric distribution will be X's distribution if X is the number of tosses you had to do in ordrer to get your first tail.

4. ## Re: Naming the distribution

Thanks, i understand why it isnt geometric now

The last part of the question says now suppose for k=1,2,.. ...Ak is the event that the first head is seen on the kth toss. Fix n>=1. Show that for all k €(1,2,.....n)

P(Ak| Xn=1)=1/n

Would I start of by writing P(Ak| Xn=1) = P(Ak ; Xn=1) / P(Xn=1)
=P(Ak n Xn=1)/ P(Xn=1)

to calculate P(Xn=1) would I plug this into the binomial formula above, then I'm not sure what to do abou P(Ak n Xn=1)?

5. ## Re: Naming the distribution

that intersection means that you had only one head in n tosses (Xn=1)
AND that head appeared in the k-th toss (Ak)
SO looking at the n independent tosses, you
${q^{n-1}p\over {n\choose 1}q^{n-1}p}={1\over n}$