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If I have $2$ independent random variables $X_{1}$ and $X_{2}$, how can I prove that $Pr (X_{1} <X_{2} <t) = \int^{t}_{-\infty} F_{X_{1}}(x) f_{X_{2}}(x) dx$

The Phenotype
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user 2000
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1 Answers1

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Observe that $$ \mathbb{P}(X_1 < X_2 < t) = \int_{-\infty}^{t} \mathbb{P}(X_2 = x, X_1 < x) \text{d}x $$ and by the independence of $X_1, X_2$ you can split the above into

$$ \int_{-\infty}^{t} \mathbb{P}(X_2 = x) \cdot \mathbb{P}(X_1 < x) \text{d}x = \int_{-\infty}^t F_{X_1}(x) f_{X_2}(x) \text{d}x $$

The first equality holds because $$ \begin{align*} \{ X_1 < X_2 < t \} &= \bigcup_{x < y < t} \{X_1 = x, X_2 = y \} = \bigcup_{y < t} \{ X_1 \leq y, X_2 = y \} \end{align*} $$

VHarisop
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  • how can I go from the $\mathbb{P}(X_{1} <X_{2} <t)$ to the integration? I mean which theorem or law can I use to do this step? – user 2000 Feb 02 '18 at 19:31
  • See edit; it is just a simple union over events, no theorems used here. – VHarisop Feb 02 '18 at 19:53
  • I am sorry, I still cannot see how the probability of the union of events will turn into an integration – user 2000 Feb 02 '18 at 20:01
  • It will be an integration since you are summing over all possible values of $y$. If $X_1, X_2$ were discrete valued, you would end up with an ordinary sum instead. – VHarisop Feb 02 '18 at 20:06
  • I still don't get it. I know that the probability of an event equals to the expectation of its indicator function, which will bring the integration to the picture. but I still cannot see how did you go from probability to integration of probability – user 2000 Feb 03 '18 at 23:56
  • user 2000: What if you have decomposed an event into a union of disjoint events? You would then have to sum the probabilities of the disjoint events to calculate the probability of the initial event, which is exactly what the integral does here. – VHarisop Feb 05 '18 at 00:42