Let there be i.i.d. r.v.s. X1,X2,…Xn with mean μ and variance σ2.
Give a value of n (a specific number) that will ensure that there is at least a 99% chance that the sample mean will be within 2 standard deviations of the true mean μ.
Let Xˉ be the sample mean, which is equal to nX1+…+Xn. Saying that there is at least a 99% chance that ∣Xˉn−μ∣<2σ is the same as saying that there is at most a 1% chance for ∣Xˉn−μ∣>2σ. Thus, we want to calculate an n such that:
P(∣Xˉn−μ∣>2σ)<0.01
P(∣Xˉn−μ∣>2σ)≤(2σ)2VarXˉn=4σ2nσ2=4n1
Therefore, if we choose n=25, we get the desired inequality.
Explain why a Gamma random variable with parameters (n,λ) is approximately Normal when n is large.
Let Xn=Y1+Y2+…+Yn. Then, we can have Xn∼Gamma(n,λ) and Yi's be i.i.d. Expo(λ). By the central limit theorem, since Xn is the sum of i.i.d. random variables, it converges to a normal distribution as n→∞.
Let Xn∼Gamma(n,λ). Determine a and b such that bXn−a→N(0,1) as n→∞
Xn∼N(λn,λ2n)
In order to convert this normal distribution to a standard normal (N(0,1)), all we need to do is subtract the mean and divide by the standard deviation. Thus:
λnXn−λn∼N(0,1)
as n→∞.
Suppose Xn is a two-state Markov chain with transition matrix
Q=(1−αβα1−β)
The rows and columns are indexed 0,1 such that q0,0=1−α,q0,1=α,q1,0=β,q1,1=1−β.
Find the stationary distribution s=(s0,s1) of Xn by solving sQ=s.
By solving sQ=s, we have that
s0=s0(1−α)+s1β and s1=s0α+s1(1−β)
s=(α+ββ,α+βα)
To verify the validity of a stationary distribution for a chain, we just need to show that siqij=sjqji, which is done if we can show that s0q01=s1q10. We have that
s0q01=α+βαβ=s1q10
Let Zn=(Xn−1,Xn). Is Zn a Markov chain? If so, what are the states and transition matrix?
Yes, Zn is a Markov Chain because conditional on Zn, Zn+1 is independent to Zn−1. This is because the components of Zn+1 and Zn−1 are either constants conditioned in Zn or independent of each other given that Xn is a Markov Chain.
The states are given as {(0,0),(0,1),(1,0),(1,1)}. The transition matrix is given as