let x0,x1,xw be iid nonegative random variables having a continuous distribtion. let n be teh first index k for which xk, xo. that is n=1. determine the probabliity mass function for n and mean e{n}.

To determine the probability mass function for n, we need to find the probability that the first index k for which xk is less than xo is equal to n. This means that x0 is the minimum value among x0, x1, ..., xn-1.

Let F(x) be the cumulative distribution function of x0. Then, the probability that x0 is less than or equal to x is F(x). The probability that all the other xi's are greater than or equal to x is (1-F(x))^(n-1), since they are all independent and identically distributed.

Therefore, the probability that n = k is the difference between the probability that x0 is less than or equal to xo and the probability that all the other xi's are greater than or equal to xo:

P(n = k) = F(xo) (1-F(xo))^(k-1) - F(xo) (1-F(xo))^k

To find the mean of n, we can use the formula for the expected value of a discrete random variable:

E{n} = Σ k P(n = k)

= Σ k [F(xo) (1-F(xo))^(k-1) - F(xo) (1-F(xo))^k]

= F(xo) Σ k (1-F(xo))^(k-1) - F(xo) Σ k (1-F(xo))^k

The first sum is an infinite geometric series with a common ratio of (1-F(xo)), so its sum is 1/(1-(1-F(xo))) = 1/F(xo). The second sum is the same series shifted by 1, so its sum is (1-F(xo))/F(xo).

Substituting these values, we get:

E{n} = 1/F(xo) - (1-F(xo))/F(xo)

= 1/F(xo) - 1 + 1/F(xo)

= 2/F(xo) - 1

Therefore, the probability mass function for n is:

P(n = k) = F(xo) (1-F(xo))^(k-1) - F(xo) (1-F(xo))^k

And the mean of n is:

E{n} = 2/F(xo) - 1

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