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I am studying some notes on exponential families and there is a section on the computation of moments. The exponential family has the form
$$exp(sum_j = 1^k phi_j B_j(x) + C(x) - D(phi))$$

I understand that for $lvert s rvert < delta$,
$$M_B(X)(s) = mathbbE[exp(s^TB(X))] < infty$$ but don't see how this implies that $mathbbE[lVert B(X) rVert ^k] < infty$. (I assume $lVert .rVert$ is the Euclidean norm)

I considered setting $s = delta_0 fracB(X)lVert B(X) rVert, delta_0 < delta$, to give

$$mathbbE[exp(s^TB(X))] = mathbbE[exp(delta_0lVert B(X) rVert]$$

but I think $s$ needs to be a constant vector and so this approach is not allowed.

I have seen the following post (Does any moment generating function implies an existence of moments?) which demonstrates how finiteness of the mgf implies that $mathbbE[lvert X rvert^k] < infty$, but cannot see how to apply it to $mathbbE[lVert B(X) rVert ^k] < infty$.

I would appreciate it if I could be shown how

$$M_B(X)(s) < infty implies mathbbE[lVert B(X) rVert ^k] < infty.$$

probability-distributions moment-generating-functions exponential-distribution

share|cite|improve this question

asked Mar 22 at 14:20

M.HarrM.Harr

12

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add a comment | 

0

$begingroup$

I am studying some notes on exponential families and there is a section on the computation of moments. The exponential family has the form
$$exp(sum_j = 1^k phi_j B_j(x) + C(x) - D(phi))$$

I understand that for $lvert s rvert < delta$,
$$M_B(X)(s) = mathbbE[exp(s^TB(X))] < infty$$ but don't see how this implies that $mathbbE[lVert B(X) rVert ^k] < infty$. (I assume $lVert .rVert$ is the Euclidean norm)

I considered setting $s = delta_0 fracB(X)lVert B(X) rVert, delta_0 < delta$, to give

$$mathbbE[exp(s^TB(X))] = mathbbE[exp(delta_0lVert B(X) rVert]$$

but I think $s$ needs to be a constant vector and so this approach is not allowed.

I have seen the following post (Does any moment generating function implies an existence of moments?) which demonstrates how finiteness of the mgf implies that $mathbbE[lvert X rvert^k] < infty$, but cannot see how to apply it to $mathbbE[lVert B(X) rVert ^k] < infty$.

I would appreciate it if I could be shown how

$$M_B(X)(s) < infty implies mathbbE[lVert B(X) rVert ^k] < infty.$$

probability-distributions moment-generating-functions exponential-distribution

share|cite|improve this question

asked Mar 22 at 14:20

M.HarrM.Harr

12

$endgroup$

add a comment | 

$begingroup$

I am studying some notes on exponential families and there is a section on the computation of moments. The exponential family has the form
$$exp(sum_j = 1^k phi_j B_j(x) + C(x) - D(phi))$$

I understand that for $lvert s rvert < delta$,
$$M_B(X)(s) = mathbbE[exp(s^TB(X))] < infty$$ but don't see how this implies that $mathbbE[lVert B(X) rVert ^k] < infty$. (I assume $lVert .rVert$ is the Euclidean norm)

I considered setting $s = delta_0 fracB(X)lVert B(X) rVert, delta_0 < delta$, to give

$$mathbbE[exp(s^TB(X))] = mathbbE[exp(delta_0lVert B(X) rVert]$$

but I think $s$ needs to be a constant vector and so this approach is not allowed.

I have seen the following post (Does any moment generating function implies an existence of moments?) which demonstrates how finiteness of the mgf implies that $mathbbE[lvert X rvert^k] < infty$, but cannot see how to apply it to $mathbbE[lVert B(X) rVert ^k] < infty$.

I would appreciate it if I could be shown how

$$M_B(X)(s) < infty implies mathbbE[lVert B(X) rVert ^k] < infty.$$

probability-distributions moment-generating-functions exponential-distribution

share|cite|improve this question

asked Mar 22 at 14:20

M.HarrM.Harr

12

$endgroup$

share|cite|improve this question

asked Mar 22 at 14:20

M.HarrM.Harr

12

share|cite|improve this question

asked Mar 22 at 14:20

M.HarrM.Harr

12

asked Mar 22 at 14:20

M.HarrM.Harr

12

asked Mar 22 at 14:20

M.HarrM.Harr

12