With K(t) = log(E[e^tX]) ,show that K'(0) = E[X]

Question

Just as the title, it is Question 83 on page 90(Ross's book, Introduction to Probability Models)

since K(t) = $log(E[e^{tX}])$

Convert $log_{10}$ to $ln_e$,

K(t) = $ln(E[e^{tX}])$/$ln$10

Then K'(t) = $E[Xe^{tX}]$ / ($E[e^{tX}]$ * $ln$10 )

K'(0) = $E[X]$ / ($E[1]$ * $ln$10)

K'(0) = $E[X]$ / $ln$10

But the solution says:

K'(t) = $E[Xe^{tX}]$ / $E[e^{tX}]$

So why is that 'log'(conversion) gone? Is that conversion necessary?

Is it possible to prove it without Measure Theory ?

Thanks in advance

QED · Accepted Answer · 2018-01-09T05:31:29.190

0

let $\mu$ be the probability measure then $$e^{K(t)}=\int e^{tx}\mathrm{d}\mu$$ Differentiating both sides w.r.t. $t$, $$e^{K(t)}K'(t)=\int\frac{d}{dt}e^{tx}~\mathrm{d}\mu=\int xe^{tx}\mathrm{d}\mu=E[Xe^{tX}]$$ So$$K'(t)=\frac{E[Xe^{tX}]}{E[e^{tX}]}$$

edited Jan 09 '18 at 05:31

answered Jan 09 '18 at 05:00

QED

12,644

Derivative of $e^{K(t)}$ is $e^{K(t)} K'(t)$, not $e^{K(t)} (1+K'(t))$. – Kavi Rama Murthy Jan 09 '18 at 05:29
@KaviRamaMurthy So sorry, I edited. A silly mistake it was, bad habit. – QED Jan 09 '18 at 05:32
@AbishankaSaha Is it possible to resolve it without Measure theory? – evergreenhomeland Jan 14 '18 at 03:06
@evergreenhomeland Just assume the random variable has a density, say $f$. Then replace $d\mu$ by $f(x)dx$ – QED Jan 14 '18 at 03:10

With K(t) = log(E[e^tX]) ,show that K'(0) = E[X]

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