Convergence in probability and variance
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> Does Xn converge 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 if Xn converges in probability, we need to evaluate lim(n→∞) P(|Xn - L| < ε), where L is the limit and ε is a small positive number.
For this problem, L can be either 1/n or n, depending on the value that Xn takes.
If Xn = 1/n, then lim(n→∞) P(|Xn - (1/n)| < ε) = lim(n→∞) P(0 < ε) = 1 for any ε > 0.
If Xn = n, then lim(n→∞) P(|Xn - n| < ε) = lim(n→∞) P(0 < ε) = 1 for any ε > 0.
Since both cases converge to 1, we can conclude that Xn converges in probability to both 1/n and n.
2> To compute E[Xn] in the limit n tends to infinity, we need to find lim(n→∞) E[Xn].
E[Xn] = Σ(x * P(Xn = x)) for all possible values of Xn.
For this problem, we have:
E[Xn] = 1/n * (1 - 1/n^2) + n * (1/n^2)
= 1/n - 1/n^3 + 1/n
= 2/n - 1/n^3
Taking the limit as n tends to infinity, we get:
lim(n→∞) E[Xn] = lim(n→∞) (2/n - 1/n^3)
= 0
Therefore, the expected value E[Xn] tends to 0 as n approaches infinity.
3> To compute Var[Xn] in the limit n tends to infinity, we need to find lim(n→∞) Var[Xn].
Var[Xn] = E[(Xn - E[Xn])^2]
Using the expression for E[Xn] obtained in part 2, we have:
Var[Xn] = E[(Xn - 0)^2]
= E[Xn^2]
For this problem, we have:
E[Xn^2] = (1/n)^2 * (1 - 1/n^2) + n^2 * (1/n^2)
= 1/n^2 - 1/n^4 + 1
= 1/n^2 + 1 - 1/n^4
Taking the limit as n tends to infinity, we get:
lim(n→∞) E[Xn^2] = lim(n→∞) (1/n^2 + 1 - 1/n^4)
= 1
Therefore, the variance Var[Xn] tends to 1 as n approaches infinity.