The random variable X has an exponential distribution with parameter b. It is found that Mx(-b^2) = 0.2
Find b
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No idea really...
Given: Mx(-b^2) = Ex(e^(-tb^2)) = 0.2
General exponential mgf: Mx(t) = b/(b-t)
Thanks !
$\displaystyle M_X(t)=\dfrac{b}{b-t}\implies M_X(-b^2)=\dfrac{b}{b-(-b^2)}=\dfrac{1}{1+b}$ 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