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Sav [38]
2 years ago
14

A random variable X has a gamma density function with parameters α= 8 and β = 2.

Mathematics
1 answer:
DerKrebs [107]2 years ago
6 0

I know you said "without making any assumptions," but this one is pretty important. Assuming you mean \alpha,\beta are shape/rate parameters (as opposed to shape/scale), the PDF of X is

f_X(x) = \dfrac{\beta^\alpha}{\Gamma(\alpha)} x^{\alpha - 1} e^{-\beta x} = \dfrac{2^8}{\Gamma(8)} x^7 e^{-2x}

if x>0, and 0 otherwise.

The MGF of X is given by

\displaystyle M_X(t) = \Bbb E\left[e^{tX}\right] = \int_{-\infty}^\infty e^{tx} f_X(x) \, dx = \frac{2^8}{\Gamma(8)} \int_0^\infty x^7 e^{(t-2) x} \, dx

Note that the integral converges only when t.

Define

I_n = \displaystyle \int_0^\infty x^n e^{(t-2)x} \, dx

Integrate by parts, with

u = x^n \implies du = nx^{n-1} \, dx

dv = e^{(t-2)x} \, dx \implies v = \dfrac1{t-2} e^{(t-2)x}

so that

\displaystyle I_n = uv\bigg|_{x=0}^{x\to\infty} - \int_0^\infty v\,du = -\frac n{t-2} \int_0^\infty x^{n-1} e^{(t-2)x} \, dx = -\frac n{t-2} I_{n-1}

Note that

I_0 = \displaystyle \int_0^\infty e^{(t-2)}x \, dx = \frac1{t-2} e^{(t-2)x} \bigg|_{x=0}^{x\to\infty} = -\frac1{t-2}

By substitution, we have

I_n = -\dfrac n{t-2} I_{n-1} = (-1)^2 \dfrac{n(n-1)}{(t-2)^2} I_{n-2} = (-1)^3 \dfrac{n(n-1)(n-2)}{(t-2)^3} I_{n-3}

and so on, down to

I_n = (-1)^n \dfrac{n!}{(t-2)^n} I_0 = (-1)^{n+1} \dfrac{n!}{(t-2)^{n+1}}

The integral of interest then evaluates to

\displaystyle I_7 = \int_0^\infty x^7 e^{(t-2) x} \, dx = (-1)^8 \frac{7!}{(t-2)^8} = \dfrac{\Gamma(8)}{(t-2)^8}

so the MGF is

\displaystyle M_X(t) = \frac{2^8}{\Gamma(8)} I_7 = \dfrac{2^8}{(t-2)^8} = \left(\dfrac2{t-2}\right)^8 = \boxed{\dfrac1{\left(1-\frac t2\right)^8}}

The first moment/expectation is given by the first derivative of M_X(t) at t=0.

\Bbb E[X] = M_x'(0) = \dfrac{8\times\frac12}{\left(1-\frac t2\right)^9}\bigg|_{t=0} = \boxed{4}

Variance is defined by

\Bbb V[X] = \Bbb E\left[(X - \Bbb E[X])^2\right] = \Bbb E[X^2] - \Bbb E[X]^2

The second moment is given by the second derivative of the MGF at t=0.

\Bbb E[X^2] = M_x''(0) = \dfrac{8\times9\times\frac1{2^2}}{\left(1-\frac t2\right)^{10}} = 18

Then the variance is

\Bbb V[X] = 18 - 4^2 = \boxed{2}

Note that the power series expansion of the MGF is rather easy to find. Its Maclaurin series is

M_X(t) = \displaystyle \sum_{k=0}^\infty \dfrac{M_X^{(k)}(0)}{k!} t^k

where M_X^{(k)}(0) is the k-derivative of the MGF evaluated at t=0. This is also the k-th moment of X.

Recall that for |t|,

\displaystyle \frac1{1-t} = \sum_{k=0}^\infty t^k

By differentiating both sides 7 times, we get

\displaystyle \frac{7!}{(1-t)^8} = \sum_{k=0}^\infty (k+1)(k+2)\cdots(k+7) t^k \implies \displaystyle \frac1{\left(1-\frac t2\right)^8} = \sum_{k=0}^\infty \frac{(k+7)!}{k!\,7!\,2^k} t^k

Then the k-th moment of X is

M_X^{(k)}(0) = \dfrac{(k+7)!}{7!\,2^k}

and we obtain the same results as before,

\Bbb E[X] = \dfrac{(k+7)!}{7!\,2^k}\bigg|_{k=1} = 4

\Bbb E[X^2] = \dfrac{(k+7)!}{7!\,2^k}\bigg|_{k=2} = 18

and the same variance follows.

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