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Let $X$ be a continuous random variable with PDF: \begin{equation} \nonumber f_X(x) = \left\{ \begin{array}{l l} 4x^3 & \quad 0 < x \leq 1\\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation} How can I calculate the moment-generating function of X and find the third moment?

What I get is MGF = $e^t * (\frac{4}{t} - \frac{12}{t^2} + \frac{24}{t^3} - \frac{24}{t^4} ) + \frac{24}{t^4}$ But, I can't get the third moment using the MGF function!!

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    $\begingroup$ Hi :) Have you calculated the powerseries of the moment generating function? Otherwise, i would guess the easiest way to get the third moment is to calculate $E(X^3)=\int_0^1 x^3\cdot f_X(x) dx$ $\endgroup$
    – Jochen
    Commented Oct 27, 2022 at 16:42

1 Answer 1

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By def: $E(e^{tX})=\int_{-\infty }^{\infty}e^{tx}f_X(x)dx=\int_{0}^{1}e^{tx}4x^3dx=\frac{(4t^3-12t^2+24t-24)e^t+24}{t^4}$

And: $\frac{\mathrm{d^3} }{\mathrm{d} t^3}E(e^{tX})=\dfrac{\left(4t^6-24t^5+120t^4-480t^3+1440t^2-2880t+2880\right)\mathrm{e}^t-2880}{t^7}=\dfrac{\left(4t^5-20t^4+80t^3-240t^2+480t-480\right)\mathrm{e}^t+\left(20t^4-80t^3+240t^2-480t+480\right)\mathrm{e}^t}{t^6}-\dfrac{6\left(\left(4t^5-20t^4+80t^3-240t^2+480t-480\right)\mathrm{e}^t+480\right)}{t^7}$
If you compute the limit at $0$ of this expression it should give you $4/7$

And by def we have: $E(X^3)=\int_{-\infty }^{\infty}x^3f_X(x)dx=\int_{0}^{1}x^34x^3dx=4/7$

So it seems to be correct.

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