Ifoa April 2012 Q7

A coin has two sides, “heads” and “tails”. Such a coin with P(heads) = p is tossed repeatedly until it lands “heads” for the first time. Let X be the number of tosses required.
Suppose the process is repeated independently a total of n times, producing values of the variables X1, X2, … , Xn , where each Xi has the same distribution as X.
Let Y = min(X1, X2, … , Xn), so Y is the smallest number of tosses required to produce a “heads” in the n repetitions of the experiment.
(i) Explain why, for each i = 1, 2, …, n, P(Xi ≥ x) is given by
P(Xi ≥ x) = (1 – p)x−1 , x = 1, 2, … . [2]
(ii) (a) Find an expression for P(Y ≥ y).
(b) Hence deduce the probability function of Y.
Please explain the solution in detail.

 

i) as \(P(X_i\geq x)\)happens, iff you get tail continuously x-1 times. (Then Head comes either on at least \(x^{th} \) toss we don't care)

ii)Y is Min of X. So if Y is not less than y, that means all the X_i are not less than y. as they are independent you can multiply each Pr.

iii)if you compare (i) and (ii),\( 1-p=(((1-p)^n))^{y-1} \) thats it.

 

the solution goes as following
1. we know that Xi's have type 1 negative binomial distribution with parameter p
\(P(X\ge x)\\=P(X=x)+P(X=x+1) +P(X=x+2).........\\=pq^{x-1}+pq^{x} + pq^{x+1} ............\\=p(q^{x-1}+q^x +q^{x+1} ..............)\) this is an infinite geometric progression with a=\(q^{x-1}\) and r=\(q\)
, thus
\(=p\frac{q^{x-1}}{1-q}\\=p\frac{q^{x-1}}{p}\\=(1-p)^{x-1}\)
2.(a)\(Y=min(X_1,X_2,......,X_n)\\thusP(Y\ge y)\\=P(min(X_1,X_2,......,X_n \ge y)\)
so if minimum of Xi's is greater than y than all Xi's should be greater than y, therefore
\(P(min(X_1,X_2,.....,X_n) \ge y)\\=P(X_1 \ge y) \times P(X_2 \ge y).........P(X_n \ge y)\\=P(X_i \ge y)^n\\=(1-p)^{n(x-1)}\)
(b) comparing this with result in part 1 we can see that Y has type 1 negative binomial distribution with probability of failure \((1-p)^n\) and probability of success \(1-(1-P)^n\)

 

sorry guys , was using latex for the first time and seems i am missing something

 

thank you , now its working