Let X1,…,Xn be i.i.d. Poisson random variables with parameter λ>0 and denote by X¯¯¯¯n their empirical average,
X¯¯¯¯n=1n∑i=1nXi.
Find two sequences (an)n≥1 and (bn)n≥1 such that an(X¯¯¯¯n−bn) converges in distribution to a standard Gaussian random variable Z∼N(0,1) .
an is sqrt(n)/sqrt(lambda)
bn is lambda
Secondly, express P(|Z|≤t) in terms of Φ(r)=P(Z≤r) for t>0 .
Write Phi(t) (with capital P) for Φ(t) .
To find sequences an and bn such that an(X¯¯¯¯n - bn) converges in distribution to a standard Gaussian random variable Z ~ N(0,1), we can use the Central Limit Theorem.
The Central Limit Theorem states that if X1, X2, ..., Xn are independent and identically distributed random variables with E[Xi] = μ and Var(Xi) = σ^2, then the normalized sum of these variables converges in distribution to a standard normal distribution as n approaches infinity.
In this case, X1, X2, ..., Xn are i.i.d. Poisson random variables with parameter λ, so E[Xi] = λ and Var(Xi) = λ.
Let's calculate the mean and variance of X¯n:
E[X¯n] = E[1/n ∑i=1^n Xi] = 1/n ∑i=1^n E[Xi] = 1/n * n * λ = λ.
Var(X¯n) = Var(1/n ∑i=1^n Xi) = 1/n^2 ∑i=1^n Var(Xi) = 1/n^2 * n * λ = λ/n.
Now, let's find an and bn such that an(X¯¯¯¯n - bn) converges to a standard normal distribution. We want the mean of X¯¯¯¯n to be 0 and the variance of X¯¯¯¯n to be 1.
Setting the mean equal to 0:
λ = 0
Solving for λ, we get:
λ = 0.
Setting the variance equal to 1:
λ/n = 1
Solving for n, we get:
n = λ.
Therefore, an = 1/√λ and bn = λ.
So, a possible choice of sequences an and bn is:
an = 1/√λ and bn = λ.
With these sequences, an(X¯¯¯¯n - bn) converges in distribution to a standard Gaussian random variable Z ~ N(0,1) as n approaches infinity.
To find the sequences (an)n≥1 and (bn)n≥1 that satisfy the condition, we can use the Central Limit Theorem (CLT). The CLT states that the sum or average of a large number of independent and identically distributed (i.i.d.) random variables will approximate a normal distribution.
In this case, we have i.i.d. Poisson random variables X1, X2, ..., Xn with parameter λ > 0. We want to find the sequences (an)n≥1 and (bn)n≥1 such that an(X¯¯¯¯n - bn) converges in distribution to a standard Gaussian random variable Z ~ N(0,1).
Step 1: Compute the mean and variance of the Poisson distribution
The mean of a Poisson random variable is given by E(Xi) = λ, which is also the variance (Var(Xi) = λ). Therefore, we have E(Xi) = Var(Xi) = λ for all i.
Step 2: Apply the Central Limit Theorem
According to the CLT, the sample mean X¯¯¯¯n = (1/n) ∑i=1n Xi follows a normal distribution with mean μ = λ and variance σ^2 = Var(Xi)/n = λ/n.
Step 3: Standardize the sample mean
To standardize the sample mean, we subtract the mean and divide by the standard deviation. In this case, we subtract μ = λ and divide by the standard deviation, which is √(σ^2) = √(λ/n).
Step 4: Rewrite the expression
Let Zn = (√n)(X¯¯¯¯n - μ)/σ, where μ = λ and σ = √(λ/n).
Step 5: Simplify the expression
Let an = √n and bn = μ = λ.
Step 6: Conclude
From the above steps, we have found the sequences (an)n≥1 = (√n) and (bn)n≥1 = λ that satisfy the condition. Therefore, an(X¯¯¯¯n - bn) converges in distribution to a standard Gaussian random variable Z ~ N(0,1).