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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.

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

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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.

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  • $\begingroup$ are you the fact that $\mathrm{Pr}(Y\geq y)=1-\mathrm{Pr}(Y\leq y)$ in the first line, second equality? $\endgroup$
    – johnny09
    Nov 3, 2019 at 2:38
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    $\begingroup$ @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 $\left( \tfrac 1 x, \,\,1\right]. \qquad$ $\endgroup$ Nov 4, 2019 at 3:27
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    $\begingroup$ @johnny09 : When you write \mathrm{Pr} instead of \Pr then you don't get proper spacing in things like this: $$ \begin{align} & \text{3\mathrm{Pr}(A): } & & 3\mathrm{Pr}(A) \\ {} \\ & \text{3\Pr(A): } & & 3\Pr(A) \\ & & & {}\quad\uparrow \\ & & & \text{different} \\ & & & \text{spacing} \end{align} $$ $\endgroup$ Nov 4, 2019 at 15:35
  • $\begingroup$ I wasn't aware of the command \Pr for writing a probability. Thanks so much for your help! $\endgroup$
    – johnny09
    Nov 4, 2019 at 15:56

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