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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 ?2012-08-30
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    Are $X$ and $Y$ supposed to be independent?2012-08-30
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    @Sasha: No, I'm looking for the joint probability of that.2012-08-30
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    @Byron: Yes, but the two events in the question I posed are implicitly dependent (I think).2012-08-30
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    @WuschelbeutelKartoffelhuhn Yes, but what about $X$ and $Y$? The question is unanswerable unless you specify how $X$ and $Y$ are related.2012-08-30
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    The way X is realized on the interval(a,b) will not impact how Y is realized (on the interval (c,d). Also, b-a is not necessarily equal to to d-c. So, independent but not necessarily identically distributed.2012-08-30
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    @WuschelbeutelKartoffelhuhn: I don't think there is a dependence issue, **if** we assume $X$ and $Y$ are independent. It comes down to an area calculation, a rectangle with an isosceles triangle bite taken out of it. Am reluctant to attempt an answer, since explaining properly (for me) requires a diagram.2012-08-30
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    @AndréNicolas: Are you afraid that readers like me might not understand your answer without you showing a diagram or is my description lacking some information/diagram?2012-08-30
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    @WuschelbeutelKartoffelhuhn: Your writeup is fine. It is just that the problem is essentially geometric, so (to me) integral is not the "right" way. By the way, you ask why the inner integral starts at $y$. It is because we want area "above" $y=x$, divided by the constant joint density. The $y$ gets "integrated out" in the second integration.2012-08-30
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    @AndréNicolas: Could you tell me where this triangle has the high point-edge (at b?) and where it converges onto the x axis (at a?)? thanks2012-08-30
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    @WuschelbeutelKartoffelhuhn: The right triangle has "legs" along $x=b$ and $y=c$. The top right-hand corner of the triangle is at $(b,b)$, bottom left is at $(c,c)$. We subtract the aarea of this from the area of a certain rectangle. Or else can find area of shape by using formula for area of a trapezoid.2012-08-30

2 Answers 2

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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

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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?2012-08-30
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    Maybe I'm misunderstanding it and the second integral ought to be inside the first.2012-08-30
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    Sorry, I disambiguated the expression, and put in parenthesis.2012-08-30
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Here is one solution without explicit integration. Let $A$ be the event that $Y>X$ and $B$ the event that $c. 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; and $A_2$ is the event that $Y>X$ and $X>c$. Then $\mathrm{Pr}(A_1)=\mathrm{Pr}(X 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.2012-09-02