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fx(x) = 2x/x^2 is a pdf of random variable X

0 < x < c

Then

Fx(x) = x^2/c^2

Find moment generating function of Y

Y = 2x+10, note it is increasing.

X = (Y-10)/2 lets call this function g(y)

Fy(y)= Fx(g(y)) = 1/(c^2)*((y-10)/2))^2

No?

And when Fy(y) is found, it gives fy(y) = (y/c^2)*(1/2-1)

Then My(t) = E[e^(ty)] = 1/(c^2) * int[e^(ty)*y(1/2-1)dy] where we integrate from (10) to (2c+10)

Where I end up with

(1/tc^2)[(e^2ct+10)*(2c+10)-e^(2ct+10t)-(10e^10t-e^10t)]

But it this the mgf of Y?

Herrlich
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    It is more convenient to find the MGF of $X$ and then to apply $M_Y(t)=\mathbb Ee^{tY}=\mathbb Ee^{t(2X+10)}=e^{10t}M_X(2t)$ – drhab Oct 10 '17 at 10:27

1 Answers1

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My(t) = E(e^(ty)) = E(e^t(2x+10)) =e^(10t)*E{e^((2t)x)}

So, My(t) = e^(10t)*Mx(2t) This can be found by simple arithmetic calculation of mgf’s.