0
$\begingroup$

I am looking at the Lemma in Ross's Stochastic Process textbook. The lemma says that

$P\{S_{N(t)} \leq s \}=\bar{F}(t)+\int_0^s \bar{F}(t-y)\mathrm{d}m(y), t \geq s \geq 0$.

In his proof, I am confused with the last step shown below:

$=\bar{F} (t)+\sum_{n=1}^{\infty} \int_0^{\infty} P\{S_n \leq s, S_{n+1} > t |S_n = y\}\mathrm{d}F_n(y)$

$= \bar{F}(t)+\sum_{n=1}^{\infty}\int_0^s \bar{F}(t-y)\mathrm{d}F_n(y)$

$=\bar{F}(t) + \int_0^s \bar{F}(t-y)\mathrm{d}\left( \sum_{n=1}^{\infty} F_n(y)\right)$

Although he mentions that the interchange of integral and summation is justified since all terms are non-negative in the bottom line, I believe it only accounts for why the summation symbol can go through the integral symbol.

But I am really confused about the last step. Can someone explain the principle behind it and answer why such step is valid?

Thanks a lot in advance.

$\endgroup$
0
$\begingroup$

For positive measures $\mu_1,\mu_2,\ldots$ it is indeed the case that $\mu:= \sum_n \mu_n$ is another measure and that $\int f d\mu = \sum_n \int f d\mu_n$ for suitable $f$ (e.g. positive, integrable). This is a measure theoretic fact and it is beyond the scope of Ross's book.

The intuitive idea is that $d(\mu_1 + \cdots + \mu_n) = d\mu_1 + \cdots + d\mu_n$ obviously for $n$ finite, and then if the integrand $f$ is (say) positive, then these increase monotonically to the infinite sum.

$\endgroup$
  • $\begingroup$ Would you mind telling me what the name of this measure theoretical fact you are referring to?? $\endgroup$ – NiubilityDiu Feb 7 '15 at 6:27
  • $\begingroup$ Link with no proof: proofwiki.org/wiki/Integral_with_respect_to_Series_of_Measures. Honestly it is obvious if you know the appropriate measure theory. $\endgroup$ – nullUser Feb 7 '15 at 6:32
  • $\begingroup$ For $f=1_A$ for a Borel set $A$, $\int f d\mu = \sum_n \int f d\mu_n$ just says $\mu(A) = \sum_n \mu_n(A)$ which is the definition. By linearity we can extend to $f=\sum a_n 1_{A_n}$ and then all positive measurable functions are increasing limits of such functions, so the monotone convergence theorem finishes the proof. $\endgroup$ – nullUser Feb 7 '15 at 6:35
  • $\begingroup$ Thank you for your clarifications. $\endgroup$ – NiubilityDiu Feb 8 '15 at 1:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.