For n>= 2, let Xn be a random variable such that P(Xn = 1/n) = 1 – 1/n^2 and P(Xn = n) = 1/n^2
1> If Xn converges in probability what is the value of the limit
Compute E[Xn] in the limit n tends to infinity and Var[Xn] in the limit n tends of infinity.
2> What is E[Xn] in the limit n tends to infinity
3> What is Var[Xn] in the limit n tends to infinity
1> To determine the value of the limit, we need to check whether Xn converges in probability.
The definition of convergence in probability states that for a sequence of random variables X1, X2, X3, ..., Xn, the limit of P(|Xn - c| >= ε) as n approaches infinity should be 0 for any ε > 0, where c is a constant.
In this case, let's take c = 0. Now, consider P(|Xn - c| >= ε):
If ε > 0, there are two scenarios:
1. If δ > 0, we have P(|Xn - 0| >= ε) = P(Xn >= ε). Since Xn can take the value of n with probability 1/n^2, we have P(Xn >= ε) = P(Xn = n) = 1/n^2.
2. If δ = 0, we have P(|Xn - 0| >= ε) = P(Xn >= ε) = 0 since Xn cannot take the value greater than n.
Therefore, for any ε > 0, P(|Xn - 0| >= ε) = 1/n^2. As n approaches infinity, P(|Xn - 0| >= ε) approaches 0.
Since the limit of P(|Xn - c| >= ε) as n approaches infinity equals 0 for any ε > 0, Xn converges in probability. The value of the limit is 0.
2> To compute E[Xn] in the limit as n tends to infinity, we can use the definition of expected value.
E[Xn] = Σx P(Xn = x), where x is the possible value of Xn.
For Xn = 1/n, the corresponding probability is P(Xn = 1/n) = 1 - 1/n^2. Therefore, E[Xn] = (1/n)(1 - 1/n^2).
As n tends to infinity, the first term (1/n) tends to 0, and the second term (1 - 1/n^2) tends to 1. Thus, the limit of E[Xn] is 0.
3> To compute Var[Xn] in the limit as n tends to infinity, we can use the definition of variance.
Var[Xn] = E[(Xn - E[Xn])^2] = E[Xn^2] - (E[Xn])^2.
For Xn = 1/n, the corresponding probability is P(Xn = 1/n) = 1 - 1/n^2. Therefore, E[Xn^2] = (1/n^2)(1 - 1/n^2).
As n tends to infinity, the first term (1/n^2) tends to 0, and the second term (1 - 1/n^2) tends to 1. Thus, the limit of E[Xn^2] is 0.
Using the result from part 2, we know that the limit of E[Xn] is 0. Therefore, the limit of (E[Xn])^2 is also 0.
Thus, Var[Xn] tends to 0 as n tends to infinity.