Independent solution

How to solve this Poisson Distribution question

Setup

Setup

Let the two Poisson means be lambda_1 and lambda_2. Their variances equal those same parameters.

V1=λ1,V2=λ2V_1=\lambda_1,\qquad V_2=\lambda_2

Model

Model

A Poisson count is zero with probability e raised to the negative mean. Express the given zero-count ratio with that identity.

e−λ1=12e−λ2e^{-\lambda_1}=\frac{1}{2}e^{-\lambda_2}

Compute

Compute

Divide the exponentials and take natural logarithms to isolate the parameter difference.

e−(λ1−λ2)=12e^{-(\lambda_1-\lambda_2)}=\frac{1}{2}
λ1−λ2=ln⁡2=0.6931471806…\lambda_1-\lambda_2=\ln 2=0.6931471806\ldots

Answer

Answer

The same difference applies to the two Poisson variances.

V1−V2=0.693(A)\boxed{V_1-V_2=0.693\quad\text{(A)}}