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I have a random variable $N(a)$, which depends on a number $a$, having the property that for all $a \in \mathbb{R}$, $$P(N(a) \geq 1) = p $$ The example I have in mind is $N(a)$ is $T-a$ where $T$ the time of first arrival in a Poisson process after $a$, which is why there is no dependence of $P(N(a) \geq 1)$ on $a$. However, let us not assume anything like this - $N(a)$ is just a random variable for each $a$.

Let $Z$ be a continuous random variable independent of all $N(a)$. I would like to assert that $$P(N(Z) \geq 1) = p.$$ My question is how I might justify this.

It is natural to try to justify it by writing

$$P(N(Z) \geq 1) = \int_{-\infty}^{+\infty} P(N(a) \geq 1) f_Z(a) ~ da = p$$ but what I don't know is how the first equality can be justified. If the variables were discrete, this would follow by conditioning, but how does one condition on the event $Z=a$ of probability $0$?

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  • $\begingroup$ "N(a) is just a random variable for each a" Actually more structure than that is needed before N(Z) can be even measurable. Which explains why the answerers, so far, have trouble addressing the core of the question. $\endgroup$ Commented Feb 2, 2015 at 3:10
  • $\begingroup$ @Did - can you explain what additional assumptions are needed and how the statement follows under these additional assumptions? $\endgroup$ Commented Feb 2, 2015 at 3:21
  • $\begingroup$ One needs at least some kind of measurability of the family $(N(a))_a$. $\endgroup$ Commented Feb 2, 2015 at 3:24
  • $\begingroup$ ...But you are not interested, since you accepted an answer relatively poor on the rigor. Allright. $\endgroup$ Commented Feb 5, 2015 at 21:49

2 Answers 2

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If $f_Z(a)>0$, then the conditional density of $N(Z)$ given $Z=a$ is $$f_{N(Z)|Z=a}(x) = \frac{f_{N(Z),Z}(x,a)}{f_Z(a)},$$ so $$\mathbb P(N(Z)\geqslant 1|Z=a) = \int_1^\infty\frac{f_{N(Z),Z}(x,a)}{f_Z(a)}\mathsf dx. $$ Note that the denominator is independent of $x$. So multiplying by $f_Z(a)$ and integrating with respect to $a$, we get $$\int_{-\infty}^\infty \mathbb P(N(Z)\geqslant 1|Z=a)f_Z(a)\mathsf da = \int_{-\infty}^\infty \int_1^\infty f_{N(Z),Z}(x,a)\mathsf dx\mathsf da. $$ Since $\mathbb P(N(Z)\geqslant 1|Z=a)=\mathbb P(N(a)\geqslant 1)$, the left-hand side is the same as the integral in your post. Using Fubini's theorem to interchange the order of integration (which is justified since the integrand is nonnegative and the integral is finite, being a probability), we have $$ \begin{align*} \int_{-\infty}^\infty \int_1^\infty f_{N(Z),Z}(x,a)\mathsf dx\mathsf da &= \int_1^\infty \int_{-\infty}^\infty f_{N(Z),Z}(x,a)\mathsf da\mathsf dx\\ &= \int_1^\infty f_{N(Z)}(x)\mathsf dx\\ &= \mathbb P(N(Z)\geqslant 1), \end{align*} $$ the desired result.

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  • $\begingroup$ I am trying to understand your answer. The second equation (i.e. $P(N(Z) \geq 1 ~|~ Z=a) = ...$). Should I understand that as a definition? Or a theorem? If it is the former, then how is the equality that it equals $P(N(a) \geq 1)$ justified? If it is the latter, then how is $P(N(Z) \geq 1 ~|~ Z=a)$ defined? $\endgroup$ Commented Feb 2, 2015 at 2:26
  • $\begingroup$ If $g$ is the density of $X$ then $\mathbb P(X\geqslant 1)=\int_1^\infty g(x)\mathsf dx$. That's all there is to that. $\endgroup$ Commented Feb 2, 2015 at 2:32
  • $\begingroup$ If you are going to write $P(N(Z) \geq 1 ~|~ Z=a)= ``\mbox{something}''$ then you must first explain how $P(N(Z) \geq 1 ~|~ Z=a)$ is defined. $\endgroup$ Commented Feb 2, 2015 at 2:45
  • $\begingroup$ ...the usual approach, I believe, is to actually use the equation you wrote as a definition of what it means to condition on an event of probability $0$. $\endgroup$ Commented Feb 2, 2015 at 2:48
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Note: $Z:\Omega\mapsto \mathbb{R} \implies N_Z \in (N_a)_{a\in \mathbb{R}}$

Therefore, we can think of $P(N_Z\geq 1)$ as another random variable $Y:=g(Z(\omega))$, so $Y$ is a function of $Z$.

Based on your definition of $N_a$, it seems like $Y=p,a.s.$, since:

$$P(Y\neq p)\equiv P(P(N_Z\geq 1)\neq p), \text{ but } \{a\in \mathbb{R}:P(N_a\geq 1)\neq p\}=\emptyset \implies P(Y\neq p)=P(\emptyset)=0$$

Therefore, $P(N_Z\geq 1)=p, a.s.$

$\square$

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