Question

Question help!
(Random variables)

Answer 1

Let X be a discrete random variable with moment generating function \(M(t)=(0.2+0.8e^t)^{10}\). Find:
a) \(P(X=5)\)
b) \(E(5^X)\)

Answer 2

I'm having trouble with a). I tried saying:
\[M(t)=E(e^{tX})=\sum_xe^{tx}f(x) \]
By the binomial theorem:
\[(0.2+0.8e^t)^{10}=\sum_{x=0}^{10}{10 \choose x}0.2^x(0.8e^t)^{10-x} \]
And then somehow pull out a \(e^{tx}\) factor, which should give me the pmf \(f(x)\), but I'm having trouble doing so... the negative in 10-x is giving me trouble.

Answer 3

Since the formula for the binomial theorem is "symmetric", I think you'll find it easier to use: \[\large (0.2+0.8e^t)^{10}=\sum_{x=0}^{10}{10 \choose x}(0.8e^t)^x0.2^{10-x}=\sum_{x=0}^{10}e^{tx}\underbrace{{10 \choose x}0.8^x0.2^{10-x}}_{f(x)} \]

So now \(P(X=5)=f(5)\), just plug in x = 5

b) \[ \large E(5^X)=\sum_{x=0}^{10}5^xf(x)=\sum_{x=0}^{10}5^x{10 \choose x}0.8^x0.2^{10-x}\\=\large \sum_{x=0}^{10}{10 \choose x}(5\cdot 0.8)^x(0.2)^{10-x}=(4+0.2)^{10}\] by the binomial theorem, and since 5(0.8) = 4

Answer 4

Sorry the 1st line got cut off, but it says:
\[ \large \sum_{x=0}^{10}e^{tx}\underbrace{{10 \choose x}0.8^x0.2^{10-x}}_{f(x)}\]

Answer 5

thank you very much!