I. Let X be a random sample from an exponential distribution with unknown rate parameter θ...

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Answer 1

Answer #1

a) P(X > 2)

b) The moment generating function of X is

$M_X(t) = E(e^{tX}) = int_{0}^{infty } e^{tx}*f_X(x)dx = int_{0}^{infty } e^{tx}* heta e^{- heta x}dx$

etrdr

$Rightarrow E(e^{tX})= int_{0}^{infty } heta e^{-( heta-t) x}dx$

(θ-t)

The mean of X is

$rac{partial }{partial t }E(e^{tX})= rac{partial }{partial t}[rac{ heta}{( heta-t)}] = rac{ heta }{( heta -t)^2}$

e-t)2

$Rightarrow E(Xe^{tX})|_{t=0}=rac{ heta }{( heta -t)^2}|_{t=0}$

$Rightarrow E(X)=rac{ heta }{ heta^2}$

E(X) =

Moreover,

$Rightarrow E(X^2e^{tX})= rac{2 heta }{( heta -t)^3}$

$Rightarrow E(X^2e^{tX})|_{t=0}= rac{2 heta }{( heta -t)^3}|_{t=0}$

20 E(X2) =

Hence, Var(X) = E(X2)-[E(X)12 = C7 92 e2 92

Var(X)2

c) Now,

The moment generating function of Y is

$M_Y(t) = E(e^{t(X_1+X_2)}) = E(e^{tX_1})*E(e^{tX_2}) = rac{ heta }{( heta -t)}* rac{ heta }{( heta -t)}$

(θ-t)

Since MGF uniquely identifies a distribution and the above mentioned MGF is a MGF of a gamma distribution, so we can write

Y ~Gamma(2,9)

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Answer 2

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