The cdf and pdf of the random variable $X(\omega)=1/\omega$

Consider the probability space $$(\Omega,\mathcal{F},\mathbb{P})$$ where $$\Omega=(0,1]$$, $$\mathcal{F}$$ is the Borel $$\sigma$$-field generated by intervals of the form $$(0,\frac{b}{2^n}]$$ with $$b\leq 2^n$$, $$b\in\mathbb{N}$$, and $$\mathbb{P}$$ is the uniform Lebesgue measure. We define the real-valued random variable $$X(\omega)=\frac{1}{\omega}$$.

I'm struggling a little bit to derive the cumulative distribution function and probability density function of $$X$$.

My attempt: $$F(x)=\mathbb{P}(\omega\in\Omega:X(\omega)\leq x)=\mathbb{P}(\omega\in\Omega:\frac{1}{\omega}\leq x)=\mathbb{P}(\omega\in\Omega:\omega\geq \frac{1}{x})$$ for $$x\in\mathbb{R}_{\geq 1}$$.

In the case $$x<1$$, we get $$\mathbb{P}(\emptyset)=0$$. That's because for small values of $$x$$, $$1/x$$ explodes but $$\omega$$ can take values up to $$1$$.

So, $$F(x)=\frac{1}{x}\mathbb{I}_{x\geq 1}$$ where $$\mathbb{I}$$ is the indicator function.

Then the probability density function is given by $$f_X(x)=\frac{d}{dx}F_X(x)=-\frac{1}{x^2}\mathbb{I}_{x\geq 1}.$$

Is my reasoning correct? I'm not sure how $$\mathcal{F}$$ plays any role here. I'd appreciate any hints.

• A check is that $F$ should be increasing: $x<y$ should imply $F(x)\le F(y)$. Does your $F$ satisfy this? Nov 3, 2019 at 1:29

If $$x\ge 1$$ then \begin{align} \Pr(X\le x) & = \Pr\left( \left\{\omega : \omega\ge \frac 1 x \right\} \right) = 1 - \frac 1 x, \\[10pt] \text{So } \frac d {dx} \Pr(X\le x) & = \frac d {dx} \, \left( 1 - \frac 1 x \right) = \frac 1 {x^2}. \end{align} You have $$1/x$$ where you needed $$1 - (1/x).$$ Note that a probability density function cannot be negative, as your proposed density function is.
• are you the fact that $\mathrm{Pr}(Y\geq y)=1-\mathrm{Pr}(Y\leq y)$ in the first line, second equality? Nov 3, 2019 at 2:38
• @johnny09 : There I was just using the fact that with the uniform distribution on $(0,1],$ the probability of being an a particular interval is just the length of the interval. The interval in this instance was $\left( \tfrac 1 x, \,\,1\right]. \qquad$ Nov 4, 2019 at 3:27
• @johnny09 : When you write \mathrm{Pr} instead of \Pr then you don't get proper spacing in things like this: \begin{align} & \text{3\mathrm{Pr}(A): } & & 3\mathrm{Pr}(A) \\ {} \\ & \text{3\Pr(A): } & & 3\Pr(A) \\ & & & {}\quad\uparrow \\ & & & \text{different} \\ & & & \text{spacing} \end{align} Nov 4, 2019 at 15:35