Question

2 Chi-square f(x)= 22)/72 2 Exponential Gamma 0<α M (t) = (1-et)" t < Normal N (μ, σ2) E (X) = μ, Var(X) = σ2

find mean and variance ,MGF of one random variable derive that step by step for number 2,3,4.Thank you

Answer 1

3:

$M_{X}(t)=E(e^{tx})=\int_{0}^{\infty}e^{tx}\cdot \frac{1}{\theta}\cdot e^{-x/\theta}dy=\frac{1}{\theta}\int_{0}^{\infty}e^{-x(1/\theta-t)}dx=\frac{1}{\theta}\left [ \frac{e^{-x(1/\theta-t)}}{1/\theta-1} \right ]_{0}^{\infty}=\frac{1}{\theta}\cdot \frac{1}{1/\theta-t}=\frac{1}{1-\theta t}$

Hence, MGF is

$M_{X}(t)=\frac{1}{1-\theta t}$

Differentiating with respect to t gives:

$M'_{Y}(t)=\frac{-1}{(1-\theta t)^{2}}\cdot (-\theta)$

The mean is:

$E(X)=M'_{X}(0)=\theta$

Differentiating with respect to t again gives:

$M''_{X}(t)=\frac{(-2)(-1)}{(1-\theta t)^{3}}\cdot (-\theta)^{2}$

So,

$E(X^{2})=M''_{X}(0)=2\theta^{2}$

The variance is:

$Var(X)=E(X^{2})-[E(X)]^{2}=\theta^{2}$

4:

Here we will use the intergral

$\int_{0}^{\infty }x^{a-1}e^{-x/b}dx=b^{a}\Gamma (a)$

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PDF of gamma distribution with parameter $\alpha >0$ and $\theta>0$ is

$f(x)=\frac{(1/\theta))^{\alpha}}{\Gamma (\alpha)}\cdot x^{\alpha-1}e^{-x/\theta},x\geq 0$

Let us assume

$b=\frac{1}{\theta},a=\alpha$ .

So pdf of gamma distribution will be

$f(x)=\frac{b^{a}}{\Gamma (a)}\cdot x^{a-1}e^{-xb},x\geq 0$

So MGF will be

$M_{X}(t)=E(e^{tx})=\int_{0}^{\infty}f(x)e^{tx}dx=\int_{0}^{\infty} e^{tx}\frac{b^{a}}{\Gamma (a)}\cdot x^{a-1}e^{-xb}dx$

ba ba 「(a) Jo

  $=\frac{b^{a}}{\Gamma (a)} \cdot \left ( \frac{1}{b-t} \right )^{a}\cdot \Gamma (a)$

  $=\left ( \frac{b}{b-t} \right )^{a}$

So MGF is

$M_{X}(t)=\left ( \frac{b}{b-t} \right )^{a}$

Now putting

$b=\frac{1}{\theta},a=\alpha$

gives

$M_{X}(t)=\left ( \frac{1}{1-\theta t} \right )^{\alpha}$
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Differentiating MGF with respect to t once gives:

Mr(t)(-a) b" abu (b - t)a+1 (b- t)a+1

Noe putting t=0 in the above equation gives:

$E(X)=M'_{X}(0)=\frac{a}{b}=\alpha \theta$

Differentiating MGF with respect to t again gives:

$M''_{X}(t)=\frac{ab^{a}(a+1)}{(b-t)^{a+2}}$

Noe putting t=0 in the above equation gives:

$E(X^{2})=M''_{X}(0)=\frac{a(a+1)}{b^{2}}$

Now putting

$b=\frac{1}{\theta},a=\alpha$

gives

$E(X^{2})=M''_{X}(0)=\alpha(\alpha+1)\theta^{2}$

Therefore variance is:

$Var(X)=E(X^{2})-[E(X)]^{2}=\alpha(\alpha+1)\theta^{2}-\alpha^{2}\theta^{2}=\alpha\theta^{2}$