Distribution function of $ X(\omega):=\frac{1}{\lambda} \ln \frac{1}{1-\omega},$ in Lebesgue measure

Question

I am having trouble in finding the right approach to this exercise:

Define the probability space $(\Omega, \mathcal{A},P) =((0,1), \mathcal{B}, \mu)$ with $\mu$ as the Lebesgue measure and $ \mathcal{B}$ the Borel-$\sigma$-algebra. Find the distribution function of the random variable $$ X(\omega):=\frac{1}{\lambda} \ln \frac{1}{1-\omega},$$ where $\lambda$ is a positive parameter.

I know that the definition of a distribution function is $P(X\leq x) $, but to be honest I'm a bit overwhelmed and don't know how to start. Does it have to do with Lebesgue integration? Because that's what we were doing last week in the lecture. Thanks in advance.

Answer 1

Let's take this one step at a time.

You have to figure out $P(X\leq x)$, the probability of the set $\{\omega\mid X(\omega)\leq x\}$, right? Well, let's try and get a feel for what set that is, first of all. It's going to be a subset of $(0, 1)$, so we're just talking about a set of real numbers here, maybe an interval or something.

So, for which $\omega$ do we have $\frac{1}{\lambda} ln \frac{1}{1-\omega}\leq x$, where $x$ is a constant in $(0, 1)$? This is just solving an inequality. At the end of it, you end up with some set $\{\omega\mid X(\omega)\leq x\}$.

Now, what is the probability of that set? Well, the probability measure in play here is just the Lebesgue measure $\mu$. You're being asked to compute $\mu(\{\omega\mid X(\omega)\leq x\})$, in other words, just the length of that set.

Answer 2

Definition $$P(X \le x) := P(\omega | X(\omega) \le x)$$

Two steps in computing $P(X \le x)$:

1. Find all $\omega$ s.t. $X(\omega) \le x$

2. Compute the probability of all those $\omega$'s.

For $x \le 0$, $P(X\leq x) = P(\emptyset) = 0$

Let $\omega \in \Omega = (0,1)$. For $x > 0$, $X(\omega) \le x$

Step 1:

$$ \iff \frac1{\lambda}\ln(\frac{1}{1-\omega}) \le x$$

$$ \iff \omega \le \frac{e^{\lambda x} - 1}{e^{\lambda x}}$$

$$ \iff \omega \in (0,1) \cap (-\infty,\frac{e^{\lambda x} - 1}{e^{\lambda x}})$$

$$ \iff \omega \in (0,\frac{e^{\lambda x} - 1}{e^{\lambda x}})$$

Step 2:

$$\mu(\omega | \omega \in (0,\frac{e^{\lambda x} - 1}{e^{\lambda x}}))$$

$$= \mu((0,\frac{e^{\lambda x} - 1}{e^{\lambda x}}))$$

$$= \frac{e^{\lambda x} - 1}{e^{\lambda x}}$$

Therefore, $F_X(x) = P(X \le x) = (1-e^{-\lambda x})1_{x \ge 0} \to f_X(x) = \lambda e^{-\lambda x}1_{x \ge 0}$, which we recognise as the exponential distribution.