Question

Let X₁, X₂, X₃ be mutually independent random variables with Poisson distributions having means 2, 1, and 4, respectively.

Find the mgf of the sum Y = X₁ + X₂ + X₃

Answer 1

The mgf of the sum Y = X₁ + X₂ + X₃, where X₁, X₂, X₃ are independent Poisson random variables is the product of their individual mgfs, resulting in M_Y(t) = e^{7(e^t-1)}.

Step-by-step explanation:

The question involves finding the moment generating function (mgf) of the sum of three independent Poisson random variables X₁, X₂, and X₃. Since they are independent, the mgf of their sum Y = X₁ + X₂ + X₃ is the product of their individual mgfs. For a Poisson distribution with mean λ, the mgf is given by M_X(t) = e^{λ(e^t-1)}.

Applying this to each X, we get:

• M_{X₁}(t) = e^{2(e^t-1)}
• M_{X₂}(t) = e^{1(e^t-1)}
• M_{X₃}(t) = e^{4(e^t-1)}

So, the mgf of Y is:
M_Y(t) = M_{X₁}(t) ‧ M_{X₂}(t) ‧ M_{X₃}(t) = e^{2(e^t-1)} ‧ e^{1(e^t-1)} ‧ e^{4(e^t-1)} = e^{(2+1+4)(e^t-1)} = e^{7(e^t-1)}.