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Original : come si procede in generale se ho delle variabili aleatorie doppie uniformi su un triangolo o su un quadrato?come determino la densità congiunta e le marginali della x e della y? c' entrano i domini normali e se c entrano come c entrano? se ho una variabile aleatoria doppia (X;Y) uniforme sul quadrato o sul triangolo come trovo la funzione di ripartizione di una variabile aleatoria U che dipende da x e da y?ho difficoltà con il seguente esercizio:trovare la distr di U=X/X+Y se (X,Y)-Unif(0,1)*(0,1) e non ben capito perchè si distinguono due casi. dove posso trovare informazioni riguardo a questi argomenti dato che sul libro di testo sono poco trattati?

Translated : How do I proceed in general if I have double uniform random variables on a triangle or on a square? How do I determine the joint density and marginals of x and y?

Do normal domains enter and if c enter how do c enter? If I have a double random variable (X; Y) uniform on the square or on the triangle as I find the distribution function of a random variable U which depends on x and y?

I have difficulties with the following exercise:

Find the distribution of $U = X / (X + Y)$ if $(X, Y) \sim \operatorname{Uniform} (0,1) \times (0,1)$. Also I don't understand well why two cases are distinguished.

Where can I find information on these topics since they are poorly treated in the textbook?

Alejandro Tolcachier
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Luca Buonacucina
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    Hey @Luca . I have tried my best to translate this question in english $($ using Google Translate $)$ . Feel free to rollback or edit as needed. – The Demonix _ Hermit Jan 08 '20 at 17:06
  • https://math.stackexchange.com/questions/2532420/probability-density-function-of-fracxxy-x-and-y-are-independently-u, https://math.stackexchange.com/questions/541322/cdf-of-x-xy-where-x-and-y-are-i-i-d-uniform – StubbornAtom Jan 09 '20 at 13:42

2 Answers2

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$$ U = \frac X {X+Y}. $$ The cumulative probability distribution function of the random variable $U$ gives the probability that (capital) $U$ is less any particular number (lower-case) $u$: $$ F_U(u) = \Pr(U\le u) = \Pr\left( \frac X {X+Y} \le u \right). $$ Since $X$ and $Y$ are positive, this fraction $X/( X+Y)$ must be between $0$ and $1.$ So we have $F_U(u) = 1$ if $u\ge1$ and $F_U(u) = 0$ if $u\le0.$ The problem is to find $F_U(u)$ when $0<u<1.$

First we can simplify the inequality $X/(X+Y) \le u$:

\begin{align} & \frac X{X+Y} \le u \\[8pt] & X \le u(X+Y) = uX+uY \\[8pt] & X-uX \le uY \\[8pt] & X(1-u) \le uY \\[8pt] & \text{Since $0<u<1,$ we have $1-u>0$} \\ & \text{so when we divide both sides by $1-u,$} \\ & \text{the “$\le$'' remains “$\le$'' and does not} \\ & \text{become “$\ge$'' :} \\[8pt] & X \le \frac u {1-u} Y. \end{align} The line $x = uY/(1-u)$ passes through the point $(0,0).$ Draw its graph and you see that it cuts the square into two parts: a triangle and a quadrilateral (except when $u=1/2,$ so that $u/(1-u)=1,$ in which case you get two isosceles right triangles.

The probability is the area of the part of the square where $x$ is smaller than it is in the other part.

Michael Hardy
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So generally you have a joint distribution of random variables $X,Y$ which is uniform over any domain $D$ in $\mathbb{R}^2$, then $$ A(D) = \text{Area}(D) = \iint_D \,dx\,dy $$ and the joint probability of $(X,Y)$ being in any region $R \subseteq D$ is given by $$ f_{XY}(R) = \frac{1}{A(D)} \iint_R \,dx\,dy $$ Deriving the individual distributions will depend on the shape of $D$, since e.g. $$ f_X(x) = \int_D f_{XY}(x,y) \,dy $$ and the bounds on the integral depend on the shape of $D$...

In your other example, you are interested in the distribution of $Z = \frac{X}{X+Y}$ where $X,Y \sim \mathcal{U}(0,1)$. Here, you can compute $$ F_Z(z) = \mathbb{P}[Z \le z] = \mathbb{P}\left[\frac{X}{X+Y} \le z\right] = \int_0^1 \int_0^1 \mathbb{I}_{\{\frac{x}{x+y} \le z\}} \,dx\,dy $$ can you absorb the indicator function into the limits of the integral?

Michael Hardy
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gt6989b
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