Rescaled Poisson random variables 2 points possible (graded) For n > 1,let Xn be a Poisson random variable with parameter 1/n . Compute P(X, 0) What can you conclude? Xn + 0 in probability, but nXn does not converge in probability X, + 0 in probability, nXn 0 in probability, and E[(nX,)?] converges Xn + 0 and nXn 7 0 in probability, but E[(nX,)?] does not converge
To compute P(Xn = 0), we can use the probability mass function (PMF) of the Poisson distribution.
The PMF of a Poisson random variable with parameter λ is given by:
P(X = k) = e^(-λ) * (λ^k) / k!
In this case, the parameter λ is 1/n, so the PMF of Xn is:
P(Xn = k) = e^(-1/n) * ((1/n)^k) / k!
Now, we want to compute P(Xn = 0), so we substitute k = 0 into the PMF:
P(Xn = 0) = e^(-1/n) * ((1/n)^0) / 0!
P(Xn = 0) = e^(-1/n) / 1
P(Xn = 0) = e^(-1/n)
Therefore, P(Xn = 0) = e^(-1/n)
Now, let's analyze the given options to see which one we can conclude:
1. Xn + 0 in probability, but nXn does not converge in probability: This is not true. Xn converges to 0 as n approaches infinity, and nXn also converges to 0 in probability since Xn converges to 0 in probability.
2. X, + 0 in probability, nXn -> 0 in probability, and E[(nXn)?] converges: This is not true, as Xn converges to 0, but nXn does not converge to 0 in probability. Also, E[(nXn)^2] does not converge.
3. Xn + 0 and nXn -> 0 in probability, but E[(nXn)?] does not converge: This is true. Xn converges to 0 as n approaches infinity, and nXn also converges to 0 in probability since Xn converges to 0 in probability. However, E[(nXn)^2] does not converge.
Therefore, the correct conclusion is:
Xn + 0 and nXn -> 0 in probability, but E[(nXn)^2] does not converge.