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According to the answer of this question a moment generating function exists if the random variable $X$ is bounded.

The proof is not quite obvious to me. More formally, let $(\Omega,\mathcal{A},\mu)$ be a probability space and $X$ be a random variable taking values in an inner product space $(K,\langle\cdot,\cdot\rangle)$ over $\mathbb{R}$.

If $X$ is bounded, then there exists a constant $M\in\mathbb{R}$, such that $||X(t)||\leq M$ for all $t\in\Omega$, where $||x|| = \sqrt{\langle x,x\rangle}$.

A moment generating function $m(t)=\mathbb{E}e^{\langle t,X\rangle}$ is said to exist, iff $$ \int_\Omega e^{\langle t,X\rangle} \, \mathrm{d}\mu < \infty. $$ Can someone point me to the proof?

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2 Answers 2

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Almost surely for $P(\mathrm d\omega)$, $\mathrm e^{\langle t,X(\omega)\rangle}\leqslant \mathrm e^{\|t\|\cdot\|X(\omega)\|}\leqslant \mathrm e^{\|t\|\cdot M}$, hence $m(t)\leqslant\mathrm e^{\|t\|\cdot M}$.

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  • $\begingroup$ Thanks. Can you show your last step in more detail? $\endgroup$ Commented Nov 6, 2014 at 8:37
  • $\begingroup$ You are welcome. Last step is that if $Y\leqslant c$ almost surely then $E(Y)\leqslant c$. $\endgroup$
    – Did
    Commented Nov 6, 2014 at 8:44
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For a fixed $t$ we have $\left\langle t,X\right\rangle \leq\left\Vert t\right\Vert \left\Vert X\right\Vert \leq\left\Vert t\right\Vert M$ and consequently $$e^{\left\langle t,X\right\rangle }\leq e^{\left\Vert t\right\Vert \left\Vert X\right\Vert }\leq e^{\left\Vert t\right\Vert M}$$


edit:

If $\mu$ is a probablity measure then for constant $c$ we find: $$\int_{\Omega}cd\mu=c\int_{\Omega}d\mu=c\mu\left(\Omega\right)=c$$ Applying that here gives: $$\int_{\Omega} e^{\left\langle t,X\right\rangle }d\mu\leq\int_{\Omega} e^{\left\Vert t\right\Vert M}d\mu=e^{\left\Vert t\right\Vert M}$$

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  • $\begingroup$ Thanks! That is the less difficult part. How to show that the integral is finite? $\endgroup$ Commented Nov 6, 2014 at 8:33
  • $\begingroup$ Then $m\left(t\right)=\int e^{\left\langle t,X\right\rangle }d\mu\leq\int e^{\left\Vert t\right\Vert M}d\mu=e^{\left\Vert t\right\Vert M}<\infty$ (as @Did remarked in his answer). $\endgroup$
    – drhab
    Commented Nov 6, 2014 at 8:36
  • $\begingroup$ Yes, but why does this last step hold? I suppose this holds only because $\mu$ is a probability measure and was not obvious to me $\endgroup$ Commented Nov 6, 2014 at 8:38
  • $\begingroup$ If $\mu$ is a probablity measure then for constant $c$ we find $\int_{\Omega}cd\mu=c\int_{\Omega}d\mu=c\mu\left(\Omega\right)=c$, since $\mu\left(\Omega\right)=1$. $\endgroup$
    – drhab
    Commented Nov 6, 2014 at 8:41
  • $\begingroup$ I mark this as correct since your last comment is the level of detail I find most useful for this question. You may wanna make it part of the answer. $\endgroup$ Commented Nov 6, 2014 at 8:49

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