# Calculating probabilities in two overlapping continuous uniform random variables (with an added constraint)

Given X~unif(a, b) and Y~unif(c, d) with a < c < b < d.

What's the probability that Y>X and Y being realized in the interval (c, b)?

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Are you asking for conditional probability $\mathsf{Pr}\left(Y>X | c < Y <b\right)$? –  Sasha Aug 30 '12 at 2:53
Are $X$ and $Y$ supposed to be independent? –  Byron Schmuland Aug 30 '12 at 2:58
@Sasha: No, I'm looking for the joint probability of that. –  Wuschelbeutel Kartoffelhuhn Aug 30 '12 at 3:04
@Byron: Yes, but the two events in the question I posed are implicitly dependent (I think). –  Wuschelbeutel Kartoffelhuhn Aug 30 '12 at 3:05
@WuschelbeutelKartoffelhuhn Yes, but what about $X$ and $Y$? The question is unanswerable unless you specify how $X$ and $Y$ are related. –  Byron Schmuland Aug 30 '12 at 3:10

Here we assume that $X$ and $Y$ are independent random variables with uniform distributions as specified by OP.

Write out the definition of the probability: $$\mathsf{Pr}\left(Y>X, c<Y<b\right) = \int_{c}^{b} \left( \int_{a}^{y}\frac{\mathrm{d}x}{b-a} \right) \frac{\mathrm{d}y}{d-c}$$ Can you finish it off?

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What is the reason for the second integral having an upper limit of y? Shouldn't it be only in terms of a, b, c, and d? –  Wuschelbeutel Kartoffelhuhn Aug 30 '12 at 3:30
Maybe I'm misunderstanding it and the second integral ought to be inside the first. –  Wuschelbeutel Kartoffelhuhn Aug 30 '12 at 3:36
Sorry, I disambiguated the expression, and put in parenthesis. –  Sasha Aug 30 '12 at 3:39

Here is one solution without explicit integration. Let $A$ be the event that $Y>X$ and $B$ the event that $c<Y<b$. Then we know that $\mathrm{Pr}(A \cap B)=\mathrm{Pr}(A|B)\mathrm{Pr}(B)$. The two probabilities on the right are simpler to calculate. First $\mathrm{Pr}(B)=\frac{b-c}{d-c}$.

Next we get a handle on $\mathrm{Pr}(A|B)$. Split the event $A$ into two disjoint events: $A_1$ and $A_2$ where $A_1$ is the event that $Y>X$ and $X<c$; and $A_2$ is the event that $Y>X$ and $X>c$. Then $\mathrm{Pr}(A_1)=\mathrm{Pr}(X<c)=\frac{c-a}{b-a}$ and $\mathrm{Pr}(A_2)=\frac{1}{2}\mathrm{Pr}(X>c)=\frac{1}{2}\left(\frac{b-c}{b-a}\right)$. Putting these two together, we have that $\mathrm{Pr}(A|B)=\frac{c-a}{b-a}+\frac{1}{2}\left(\frac{b-c}{b-a}\right)$.

Finally, using $\mathrm{Pr}(A \cap B)=\mathrm{Pr}(A|B)\mathrm{Pr}(B)$ and doing some algebra we find that $\mathrm{Pr}(A \cap B)=\frac{-2ab+b^2+2ac-c^2}{2(a-b)(c-d)}$.

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Two things are unclear for me here: In Pr(A_2) where did the factor 1/2 come from and how was Y>X taken into account? Maybe the question answers itself, but I don't see how. –  Wuschelbeutel Kartoffelhuhn Sep 2 '12 at 4:51