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?
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 check if Xn converges in probability, we need to compute the limit of the probability of Xn being equal to a particular value, as n tends to infinity.
First, let's find the limit of P(Xn = 1/n) as n tends to infinity. Since 1/n^2 -> 0 as n tends to infinity, the limit of P(Xn = 1/n) is 1.
Next, let's find the limit of P(Xn = n) as n tends to infinity. Since 1/n^2 -> 0 as n tends to infinity, the limit of P(Xn = n) is 0.
Therefore, the probability distribution of Xn converges to the constant random variable X, where P(X = 0) = 1 and P(X = x) = 0 for all x ≠ 0.
So, Xn converges in probability to the constant random variable X.
2> To compute the expectation E[Xn], we need to find the limit of E[Xn] as n tends to infinity.
E[Xn] = Σ(x * P(Xn = x))
For Xn = 1/n, E[Xn] = (1/n) * (1 - 1/n^2)
= 1/n - 1/n^3
For Xn = n, E[Xn] = n * (1/n^2)
= 1/n
In the limit as n tends to infinity, both terms approach 0.
Therefore, E[Xn] in the limit n tends to infinity is 0.
3> To compute the variance Var[Xn], we need to find the limit of Var[Xn] as n tends to infinity.
Var[Xn] = E[(Xn - E[Xn])^2]
= E[(Xn - 0)^2]
= E[Xn^2]
For Xn = 1/n, E[Xn^2] = (1/n^2) * (1 - 1/n^2)
= 1/n^2 - 1/n^4
For Xn = n, E[Xn^2] = n^2 * (1/n^2)
= 1
In the limit as n tends to infinity, the first term approaches 0 and the second term remains 1.
Therefore, Var[Xn] in the limit n tends to infinity is 1.