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$f(x) = 2x/c^2$, $0<x<c$

Integration of $2xe^{tx}/c^2$ between 0 and c gives the MGF;

$2e^{tc}/c$.

But it does not seem right. I cant use it to find expected value. What is wrong?

Guy Fsone
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Herrlich
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1 Answers1

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This is a triangular distribution with pameters $a=0$ and $b=c$.

By direct integration $$ M_X(t)=\Bbb E(\mathrm e^{tX})=\frac{2}{c^2}\int_0^c x\,\mathrm e^{tx}\mathrm dx=\tfrac{2}{c^2}\frac{\mathrm d}{\mathrm dt}\left(\int_0^c \mathrm e^{tx}\mathrm dx\right)=\tfrac{2}{c^2}\frac{\mathrm d}{\mathrm dt}\left(\tfrac{\mathrm e^{ct}-1}{t}\right)=2\frac{\mathrm e^{ct}(ct-1)+1}{t^2c^2} $$ and the Taylor expansion using the expansion for $\mathrm e^{ct}$ is $$ M_X(t)=1+\frac{2c}{3}t+\frac{c^2}{2}\frac{t^2}{2}+o(t^3)=1+M_X'(0)t+M''_X(0)\frac{t^2}{2}+o(t^3) $$ and then $\Bbb E(X)=\frac{2c}{3}$

alexjo
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  • What's the MGF, and could you clarify how to calculate the E(X) from it. Is THE MGF I came up with wrong? – Herrlich Oct 06 '17 at 12:34
  • Yes you MGF is wrong. I've added the detail for the mean. – alexjo Oct 06 '17 at 12:42
  • Are you missing an "x" inside the integration at second row of your calc, or am I wrong? – Herrlich Oct 06 '17 at 12:55
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    Do you mean with the $t$-derivative? – alexjo Oct 06 '17 at 12:57
  • Yes exactly. That row. – Herrlich Oct 06 '17 at 13:47
  • Observe that I've used the Leibniz integral rule for differentiation under the integral sign $$\int_0^c \Big(x,\mathrm e^{tx}\Big)\mathrm dx=\int_0^c \frac{\mathrm d}{\mathrm dt}\left(\mathrm e^{tx}\right)\mathrm dx=\frac{\mathrm d}{\mathrm dt}\left(\int_0^c \mathrm e^{tx}\mathrm dx\right)=\tfrac{2}{c^2}\frac{\mathrm d}{\mathrm dt}\left(\tfrac{\mathrm e^{ct}-1}{t}\right)$$ – alexjo Oct 06 '17 at 13:54
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    It’s very much possible to do it with integration by parts. The use of that theorem seems sneaky if you aren’t used to it. – DaveNine Oct 06 '17 at 23:02