Can I make the substitution of dP when using the CDF?

Question

Random variable $X \geq 0$ with parameter $\lambda>0$ and $X$ has the c.d.f.

$$ F (a) = P{(X ≤ a)} = 1 − \exp(−λa)$$ for $a \geq 0$.

Consider $Z = (λ'/λ)\exp(-(λ'-λ)X)$

Show that $E[Z]=1$ thus I want to evaluate $\int_{(-\infty ; \infty)}Z(w) \, dP(w)$

1) Am I allowed to differentiate the cumulative distribution function $F(a)$ to obtain $dP$ :

$$dF = dP = λ\exp(-λx) \, dx$$

then make a substitution into the integral. If yes , is that because of the fundamental theorem of calculus ? I think this also will change the integral from $(-\infty ; \infty)$ to $(0,\infty)$

2) Further more, how can you relate the $dP(w)$ which is the measure for the event w to the $λ\exp(-λx) \, dx$ ? i.e the intuition or explanation.

Answer 1

1. You are right. Actually the cdf is $F(x) = (1-\exp(-\lambda x))_+$ whose derivative (in terms of distributions) is $\lambda \exp(-\lambda x)1_{x\ge 0}$, this is where the change of interval of integration comes from.
2. You can roughlt say that $dP_X(x) = \lambda \exp(-\lambda x)1_{x\ge 0}dx$ is the probability $P(X\in [x,x+dx])$. This is the interpretation of $dP_X$.