Let X be a Binomial random variable with parameters n and 0 < p < 1.
Show that lim E(1+X)·E(1+X)·E(1/1+X)=1.
n→∞
To prove the limit expression, we'll need to make use of the properties of expectation and the properties of limits.
Step 1: Expand the expression.
Let's start by simplifying the expression E(1+X)·E(1+X)·E(1/1+X):
E(1+X)·E(1+X)·E(1/1+X) = [E(1+X)]²·E(1/1+X)
Step 2: Express the expectations in terms of the random variable X.
The expected value of 1+X can be expressed as:
E(1+X) = 1 + E(X), where E(X) is the expected value of X.
Similarly, the expected value of 1/1+X can be expressed as:
E(1/1+X) = 1/(1+E(X)).
Substituting these expressions back into the expanded form, we have:
[E(1+X)]²·E(1/1+X) = [(1 + E(X))²]·(1/(1+E(X)))
Step 3: Simplify the expression.
Expanding the squared term, we get:
[(1 + E(X))²] = 1 + 2E(X) + [E(X)]²
Simplifying further, we have:
[(1 + E(X))²]·(1/(1+E(X))) = [1 + 2E(X) + [E(X)]²]·(1/(1+E(X)))
Canceling out the terms (1+E(X)), we obtain:
[1 + 2E(X) + [E(X)]²]·(1/(1+E(X))) = 1 + 2E(X) + [E(X)]²
Step 4: Take the limit as n approaches infinity.
We need to show that the expression 1 + 2E(X) + [E(X)]² approaches 1 as n approaches infinity.
Since we are given that X is a binomial random variable with parameters n and 0 < p < 1, we know that the expected value of X is given by E(X) = np.
Taking the limit as n approaches infinity, we have:
lim(n→∞) [1 + 2E(X) + [E(X)]²] = 1 + 2(np) + (np)²
Since p is less than 1, (np)² will approach 0 as n approaches infinity. Likewise, 2(np) will also approach 0.
Therefore, the expression simplifies to:
lim(n→∞) [1 + 2E(X) + [E(X)]²] = 1
Thus, we have shown that lim E(1+X)·E(1+X)·E(1/1+X) = 1 as n approaches infinity.