0

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.

Community
  • 1
NiubilityDiu
  • 70

1 Answers1

0

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.

nullUser
  • 28,703
  • Would you mind telling me what the name of this measure theoretical fact you are referring to?? – NiubilityDiu Feb 07 '15 at 06:27
  • Link with no proof: https://proofwiki.org/wiki/Integral_with_respect_to_Series_of_Measures. Honestly it is obvious if you know the appropriate measure theory. – nullUser Feb 07 '15 at 06:32
  • 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. – nullUser Feb 07 '15 at 06:35
  • Thank you for your clarifications. – NiubilityDiu Feb 08 '15 at 01:56