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For continuous random variable X,

pdf: $f_{X}(x)=2(1-x), x\in[0,1]$

mgf: $M_{X}(t)=\frac{2(e^t-t-1)}{t^2}$

Problem is to find mean and variance from mgf, I tried using $\frac{d}{dt}M_{X}(0)$ and $\frac{d}{dt}[ln(M_{X}(0))]$. But I can't seem to solve it even if I use L'Hôpital's rule and $\lim_{t\to 0}\frac{e^t-1}{t}=1$. How can the mean and variance be found using the mgf only?

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For $|t| <1$ we have $M_X(t)=2(\frac 1 2+t/3!+t^{2}/4!+...)$. It is easy to write down $M_X'(0)$ amd $M_X''(0)$ from this. I will let you finish the computation.

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  • $\begingroup$ Thanks, this really answered my question, but how did you get $𝑀_{𝑋}(𝑡)=2(1/2+t/3!+t^2/4!+...)$ ? $\endgroup$ – dylan May 19 at 5:26
  • $\begingroup$ From the power series for $e^{t}$.@dylan $\endgroup$ – Kavi Rama Murthy May 19 at 5:28

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