Let N be a geometric r.v. with mean 1/p; let A1,A2,… be a sequence of i.i.d. random variables, all independent of N, with mean 1 and variance 1; let B1,B2,… be another sequence of i.i.d. random variable, all independent of N and of A1,A2,…, also with mean 1 and variance 1. Let A=∑Ni=1Ai and B=∑Ni=1Bi.
Find the following expectations using the law of iterated expectations. Express each answer in terms of p using standard notation.
E[AB]=- unanswered
E[NA]=- unanswered
Let N^=c1A+c2 be the LLMS estimator of N given A. Find c1 and c2 in terms of p.
c1= - unanswered
c2=1 - unanswered
1
Can someone please answer?
yes please someone who loves math¡
c2 = 1
To find the expectations using the law of iterated expectations, we need to know the conditional expectations of the random variables involved. Let's start by finding E[AB].
Using the law of iterated expectations, we can express E[AB] as:
E[AB] = E[E[AB|N]]
Now, let's find E[AB|N]. Since A and B are independent of N, we can express it as:
E[AB|N] = E[A|N] * E[B|N]
To find E[A|N], we can use the fact that A is a sum of independent and identically distributed random variables A1, A2, ..., AN:
E[A|N] = E[A1 + A2 + ... + AN|N]
= E[A1|N] + E[A2|N] + ... + E[AN|N]
= NE[A1|N]
= NE[A1]
= N
Similarly, E[B|N] = N
Substituting these values back into the expression for E[AB|N], we get:
E[AB|N] = N * N
= N^2
Finally, substituting this expression into the expression for E[AB], we get:
E[AB] = E[E[AB|N]]
= E[N^2]
To find E[N^2], we need to know the distribution of N. Since N is a geometric random variable with mean 1/p, its variance can be calculated as (1 - p)/(p^2). Therefore, we can express E[N^2] as:
E[N^2] = Var(N) + [E[N]]^2
= (1 - p)/(p^2) + (1/p)^2
= (1 - p + 1/p^2)
So the answer to E[AB] is (1 - p + 1/p^2).
Now, let's move on to finding E[NA].
Similarly, using the law of iterated expectations, we can express E[NA] as:
E[NA] = E[E[NA|N]]
To find E[NA|N], we can express it as:
E[NA|N] = E[N * A|N]
= N * E[A|N]
= N^2
Therefore, substituting this expression back into the expression for E[NA], we get:
E[NA] = E[E[NA|N]]
= E[N^2]
Using the same approach as before, we know that E[N^2] is (1 - p + 1/p^2).
Finally, let's find c1 and c2 in terms of p for the LLMS estimator N^ = c1A + c2.
To find c1, we can use the fact that N^ is the conditional expectation of N given A. Therefore, c1 can be calculated as:
c1 = Cov(N, A) / Var(A)
Since A and N are independent, their covariance is 0. Therefore, we have:
c1 = 0 / Var(A)
= 0
To find c2, we can express N^ as the conditional expectation of N given A:
N^ = E[N|A]
= E[N]
= 1/p
Therefore, c2 = 1.
In summary:
E[AB] = 1 - p + 1/p^2
E[NA] = 1 - p + 1/p^2
c1 = 0
c2 = 1
E[AB] and E[NA] are both(2-p)/(p^2).
Let N^=c1A+c2 be the LLMS estimator of N given A. Find c1 and c2 in terms of p.
c1= 1-p
PLEASE, could you help out by giving away the answer for c2 ???