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I know this is quite basic, but I fail to see where my mistake with the following simple example from wikipedia is.

\begin{align*} G(y) &= \Pr(Y \leq y) \\ &= \Pr \left (X \geq \frac{1}{y} \right) \\ &= 1 - \Pr\left(X < \frac{1}{y} \right) \\ &= 1 - F\left(\frac{1}{y} \right) \end{align*}

What I don't understand is, why does first CDF, i.e. $G(y) = \Pr(Y \leq y)$, say "smaller or equal", while the second CDF, i.e. $F\left(\frac{1}{y} \right)= \Pr\left(X < \frac{1}{y} \right)$, says "strictly smaller"?

Shouldn't they both have the same definition, i.e. "smaller or equal", as they're both just a normal CDF? Can you please explain to me what I'm missing?

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    $\begingroup$ It is assumed that $X$ is a continuous random variable with a density and so $\Pr(X=x) = 0$ for all $x$. It follows that $\Pr\left(X < \frac{1}{y} \right) = \Pr\left(X \leq \frac{1}{y} \right) = F\left(\frac{1}{y} \right)$ $\endgroup$
    – periwinkle
    Commented Oct 29, 2021 at 14:15
  • $\begingroup$ I see, thanks a lot, that makes a lot of sense! $\endgroup$
    – futur3boy
    Commented Oct 29, 2021 at 14:20

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You have

$$ \Pr\Big(X \ge \frac{1}{y} \Big) = 1 - \Pr\Big(X < \frac{1}{y} \Big) $$

it would be contradictory if you had "$=$" on both sides of the equation. Moreover, as noticed in the comment, for continuous random variable $\Pr(X = x) = 0$ so it doesn't really matter.

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