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

Answer 1

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