The random variable X has an exponential distribution with parameter b. It is found that Mx(-b^2) = 0.2

Find b

****************

No idea really...

Given: Mx(-b^2) = Ex(e^(-tb^2)) = 0.2

General exponential mgf: Mx(t) = b/(b-t)

Thanks !

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- Sep 16th 2012, 08:09 PMTheCountadorMoments
The random variable X has an exponential distribution with parameter b. It is found that Mx(-b^2) = 0.2

Find b

****************

No idea really...

Given: Mx(-b^2) = Ex(e^(-tb^2)) = 0.2

General exponential mgf: Mx(t) = b/(b-t)

Thanks ! - Sep 17th 2012, 08:36 AMharish21Re: Moments
$\displaystyle M_X(t)=\dfrac{b}{b-t}\implies M_X(-b^2)=\dfrac{b}{b-(-b^2)}=\dfrac{1}{1+b}$ and you can use the given condition to find b.

If you still have trouble, try using the defintion of mgf

$\displaystyle M_X(-b^2)=\int_0^{\infty}\exp(-b^2x)\times b \times \exp(-bx)\;dx$

and integrate - Sep 17th 2012, 05:11 PMTheCountadorRe: Moments
Thank you ! I should have known that :p