Answer to Question #283678 in Statistics and Probability for Ahmed Hamza

Question

Let 𝑀𝑋 (𝑡) = 𝑘𝑒 𝑡 1−0.8𝑒 𝑡 with µ=5 and define 𝑍 = 2𝑋 +3 , find Var(Z).

Answer 1

Solution:

$M_X(t)=ke^{t\left(1-0.8e^t\right)}, \mu=5,Z=2X+3$

$\Rightarrow M'_X(t)=ke^{t\left(1-0.8e^t\right)}\left(1-0.8e^t-0.8e^tt\right) \\E(X)=\mu=5\ (given) \\E(X)=M'_X(0)$

So, $k(1)\left(1-0.8(1)-0.8(1)(0)\right)=5$

$\Rightarrow k(0.2)=5 \\ \Rightarrow k=25$

So, we have, $M_X(t)=25e^{t\left(1-0.8e^t\right)}$

Now, $M''_X(t)=25[0.64e^{3t-0.8e^tt}t^2-2.4e^{2t-0.8e^tt}t+1.28e^{3t-0.8e^tt}t-3.2e^{2t-0.8e^tt}+0.64e^{3t-0.8e^tt}+e^{t\left(-0.8e^t+1\right)}]$

$M''_X(0)=E(X^2)$

So, $M''_X(0)=25[0.64(0)-2.4(0)+1.28(0)+3.2(1)+0.64(1)+1]$

$=25[3.2+1.64]=121$

$Var(Z)=Var(2X+3)=4Var(X)+0=4Var(X)$

Also, $Var(X)=E(X^2)-[E(X)]^2=121-5^2=96$

Then, $Var(Z)=4(96)=384$