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

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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}
----------------

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}

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