Let X be a Binomial random variable with parameters n and 0 < p < 1.
Show that lim E(1 + X) · E(1/1 + X)= 1.
n→∞
To show that lim E(1 + X) · E(1/1 + X) = 1 as n approaches infinity, we need to compute the expected values and take the limit.
Let's start by computing E(1 + X) and E(1/1 + X):
E(1 + X) = ∑(k=0 to n) (1 + k) * nCk * p^k * (1 - p)^(n-k)
To simplify the expression, expand (1 + k) as (k+1) * nCk = (k+1) * n!/((n-k)! * k!).
E(1 + X) = ∑(k=0 to n) (k+1) * nCk * p^k * (1 - p)^(n-k)
Next, let's compute E(1/1 + X):
E(1/1 + X) = ∑(k=0 to n) 1/(1+k) * nCk * p^k * (1 - p)^(n-k)
Now, let's denote the two expressions as A and B:
A = E(1 + X) and B = E(1/1 + X).
We will take the limit of A * B as n approaches infinity:
lim (n→∞) A * B = lim (n→∞) [∑(k=0 to n) (k+1) * nCk * p^k * (1 - p)^(n-k)] * [∑(k=0 to n) 1/(1+k) * nCk * p^k * (1 - p)^(n-k)]
Since we have n in both the summations, we can rewrite them using the properties of the binomial coefficient:
lim (n→∞) A * B = lim (n→∞) [∑(k=0 to n) (k+1) * (n! / ((n-k)! * k!)) * p^k * (1 - p)^(n-k)] * [∑(k=0 to n) 1/(1+k) * (n! / ((n-k)! * k!)) * p^k * (1 - p)^(n-k)]
Now, we can multiply the two summations and simplify the expression:
lim (n→∞) A * B = lim (n→∞) ∑(k=0 to n) [(k+1)/(1+k)] * [n!^2 / ((n-k)!^2 * k!^2)] * [p^(2k) * (1 - p)^(2n-2k)]
Notice that the term [(k+1)/(1+k)] simplifies to 1, and we can cancel out the n!^2 term.
lim (n→∞) A * B = lim (n→∞) ∑(k=0 to n) [p^(2k) * (1 - p)^(2n-2k)]
To evaluate the limit, we need to compare the exponents of p and (1-p) to the square of n. Upon comparison, we can see that as n approaches infinity, the exponents get dominated by 2n.
The term p^(2k) approaches zero as n approaches infinity because p < 1, and the term (1-p)^(2n-2k) approaches zero as n approaches infinity because 1-p > 0.
Therefore, all the terms in the summation become negligible, except when k is very close to n.
lim (n→∞) A * B = lim (n→∞) [p^(2n) * (1 - p)^(2n)]
Since p > 0 and p < 1, we can rewrite the expression as:
lim (n→∞) A * B = 0 * 0 = 0
Hence, we have shown that lim E(1 + X) · E(1/1 + X) = 0, not equal to 1.
Note that there might be an error in the derivation or assumption made, as the statement initially provided (lim E(1 + X) · E(1/1 + X) = 1) does not hold true.