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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.

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  • $\begingroup$ Where did you get this please? This seems to be precisely what I'm looking for here Dumb question: Computing expectation without change of variable formula $\endgroup$ – BCLC Apr 28 '18 at 12:00
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    $\begingroup$ It's one of my exercises for my probability theory I lecture. So it's pretty random that it's helping you. But still nice to hear. $\endgroup$ – d237 Apr 29 '18 at 8:57
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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.

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  • $\begingroup$ Thanks for your answer. So i gotta use the Lebesgue measure. But how can i consider the whole interval (0,1) for x? I can see the inequality, but for the distribution function don't I have to view every case in this uncountable infinite interval? $\endgroup$ – d237 Apr 26 '18 at 20:11
  • $\begingroup$ @d237 Yes, determining the distribution function means computing $P(X<x)$ for every $x\in \mathbb R$. But that's not a problem. Just let $x$ be a constant like you would in any algebra problem. If I solve $ax+b=0$, I have to prove my answer works "for all $a,b$", but that doesn't mean I have to do an uncountably infinite amount of work. $\endgroup$ – Jack M Apr 26 '18 at 20:18
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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.

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