Question

Suppose N has a geometric distribution with parameter p Derive a closed form expression for E N I N <= k), k = 1,2 . Check via simulaton for p = 0.2, k = 3

Answer 1

The geometric distribution is $P(N=n)=(1-p)^{n-1}p;n=1,2,3,...$

The probability

n-1 n=1 k) = 1-(1-02)3 P(N

Now, the sum

n-1 dq n-1 dq


k+1 dq (1-q)2 n=1 n=1

The conditional expectation,

1-(1 -P) -P1-P)

For .

k = 3,p 0.2

0.2 + 2 x 0.2 x 0.83 x 0.2 x 0.82 0.488 E(NIN < 3) 1.852459

The above is the theoretical expectation. The R code for finding the expectation via simulation is given below.

n <- 10000
k <- 3
N <- rgeom(n,prob=0.2)
N <- N+1
N_k <- N[N<=k]
mean(N_k)

The simulated expectation is 1.855763.