expected value ques

[Sinistral]

is the ans -1/2? I am not comfortable with my solution.

[Sinistral]

I am getting E[P] = 100+n (where n is the total number of fish in the pond before 100 tagged fish are put in)
will it be unbiased? or am I missing something and E[P] should come out to be n (the true population of the fish discounting the tagged fishes)??

[Amit Goyal]

E(N-1) cannot be -1/2 since N-1 always takes non-negative values.
Note that
E(N -1) = E(N) - 1 (by linearity of expectation)
Now let us find E(N) using
E(N) = E(E(N|X))

E(N|X = x) = E(inf{n | Y(n) > x}|X  = x)
               = E(inf{n | Y(n) > x}) (Since X and Y(n) are independent draws)
Note that for 0 < x < 1,
inf{n | Y(n) > x} has an interpretation: the number of trials until first success, where success is defined as Y(n) > x. And the probability of success in each independent trial is (1-x).  
E(N|X = x)
= E(inf{n | Y(n) > x})
= 1(1-x) + 2x(1-x) + 3x^2(1-x) +......
= 1/(1-x)

Now coming back to E(N):
E(N) = E(E(N|X)) = E(1/(1-X)) = ∞ (infinity)
Thus,
E(N-1) = ∞.

[Amit Goyal]

Let M be the number of fishes in the pond before 100 tagged fishes were released. Probability of catching a tagged fish in each trial will then become 100/(100+M). Number of fishes (N) you catch till you get the tagged one is a geometric random variable with probability of success in each trial equal to 100/(100+M). Thus,
E(N) = (100+M)/100. M is estimated using P = 100 x N. For P to be an unbiased estimator of M, it must be the case that E(P) = M. Clearly,
E(P)
= 100(E(N))
= 100 + M
≠ M
Thus, the estimator P is biased.
V(P)
= 10000(V(N))
= M(100+M)

[Sinistral]

Amit Goyal wrote

inf{n | Y(n) > x} has an interpretation: the number of trials until first success, where success is defined as Y(n) > x. And the probability of success in each independent trial is (1-x).  

why is the probability of success in each independent trial (1-x)?

and yes thank u so much :)

[Sinistral]

oh got it!! :)

thanx again :)

[anon_econ]

I didn't get how the probability of catching a tagged fish in each trial is the same. I thought P(N=2) would be (M/100+M)*(100/99+M). Are we approximating it by an infinite population?

[anon_econ]

Also, Sinistral, would you mind sharing with us where you got this question from?

[Amit Goyal]

It is given in the question: You keep catching and releasing the fish until you get the tagged one.

[anon_econ]

Oops. I missed that. Thanks.

[Sumit]

Hey Sinstral,
These questions are from DSE option-B paper..Isn't it????

[Sinistral]

2004 Option B Ques 8(B)

[anon_econ]

Thanks!