$(X,Y)$ is a two dimensional random variable that is uniformly distributed in the triangle : $(0,0),(1,a),(1,b)$ where $0 Find $f_{X,Y}$(s,t). (The Joint probability distribution of $(X,Y)$). The solution is: Let $T$ be the set of the points in $\mathbb{R}^2$ inside that triangle,and let $S$ be the area of that triangle. So we know that $S= \frac{b-a}{2}$,and that's why $f_{X,Y}(s,t) = \frac{2}{b-a}\chi_T(s,t)$ where $\chi_T(s,t)$ is an indicator that the point $(s,t)$ is in $T$. But I don't understand the solution.
Probability distribution for a uniform random point in triangle
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probability
probability-theory
probability-distributions
random-variables
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0How can we understand why you "don't understand,why is that the solution" if you do not say? – 2017-01-27
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0@Did I don't understand why $1/S$ would be the Joint probability distribution? – 2017-01-27
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0Nobody said that 1/S is the joint probability distribution, or any distribution for that matter. 1/S is a **number**, right? – 2017-01-27
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0@Did Yes,it is a number.So how come thats the Joint probability distribution? – 2017-01-27
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0It is not. (Bis.) – 2017-01-27
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0@Did Could you please,explain to me why $f_{X,Y}(s,t)=\frac{2}{b-a}X_T(s,t)$ the Joint probability distribution? – 2017-01-27
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0Sure, as soon as you explain how you think one should compute it... – 2017-01-27
1 Answers
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I assume that you understand that $S=(b-a)/2$.
Uniform distribution means that the probability density function has the same value everywhere in the triangle, say $f_{X,Y}(s,t)=c$ whenever $(s,t)$ is in the triangle -- i.e., $\chi_T(s,t)=1$. You also know that "probabilities add-up to 1", which in this case translate as: $$ \iint f_{X,Y}(s,t)\, ds\, dt=\iint c\, \chi_T(s,t)\,ds\,dt= \int_S c\,ds\,dt=c\,\int_S ds\,dt = c\,S= 1. $$ Therefore $c=1/S$ and the result follows.