Applying Linear Functions to a Random Sequence
3 points possible (graded)
Let (Zn)n≥1 be a sequence of random variables such that
n−−√(Zn−θ)−→−−n→∞(d)Z
for some θ∈R and some random variable Z.
Let g(x)=5x and define another sequence by Yn=g(Zn).
The sequence n−−√(Yn−g(θ)) converges. In terms of Z, what random variable does it converge to?
n−−√(Yn−g(θ))−→−−n→∞(d)Y.
(Answer in terms of Z)
Y=
To find the random variable that the sequence n−−√(Yn−g(θ)) converges to in terms of Z, we need to first substitute the expression for Yn and g(θ).
Yn = g(Zn)
Yn = 5Zn
Now we can rewrite the sequence as:
n−−√(Yn−g(θ))−→−−n→∞(d)Yn
n−−√(5Zn−g(θ))−→−−n→∞(d)Yn
Since g(θ) = 5θ, we can substitute it into the expression:
n−−√(5Zn−5θ)−→−−n→∞(d)Yn
Now, we can simplify the expression:
n−−√(5(Zn−θ))−→−−n→∞(d)Yn
Since n−−√(Zn−θ)−→−−n→∞(d)Z, we can substitute it in:
n−−√(5(Zn−θ))−→−−n→∞(d)5Z
Therefore, the random variable that the sequence n−−√(Yn−g(θ)) converges to in terms of Z is 5Z.