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I have two probability density functions where i need to find the distribution function.

The first function is $$f(x,y)= \begin{cases} \frac{x}{y} & \text{for $0\leq x\leq y\leq c$}\\ 0&\text{otherwise}\\ \end{cases}$$

$c$ has to be found so that this is a probability density. I am aware that I have to integrate the function twice with $c$ as the upper limit so that the distribution function is $1$ at is upper limit. Am I correct to assume that I can solve for $c$ with $$F (x,y) = 1 = \int_{0}^{c} \int_{0}^{y} \frac{x}{y} \, \mathrm{d}y \, \mathrm{d}x$$

The second density function is $$f(x,y)= \begin{cases} \frac{x}{y} & \text{for $0\leq x\leq y\leq 2x$}\\ 0&\text{otherwise}\\ \end{cases}$$

Here, me and my classmates are struggling to interpret any idea into the boundaries of the integer for the probability distribution function. Has anyone an idea how the boundaries are formed in such a case?

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  • $\begingroup$ The second example doesn't put an explicit upper boundary for $x$ so you need to put in a value for the upper boundary for $x$ that makes $f(x,y)$ integrate to 1. The relation $0\leq x \leq y \leq 2x$ says that $x>0$ and, if so, $x\leq 2x$. Then just do what you did for the first example. $\endgroup$
    – JimB
    Commented May 28 at 5:04

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Am i correct to assume that i can solve for c with $$F (x,y) = 1 = \int_{0}^{c} \int_{0}^{y} \frac{x}{y} \ dydx$$

No. Your underlying principle is correct, in that you should set $$ 1 = \iint\limits_{{0 \le x \le y \le c}} \frac x y \, \mathrm{d}A $$ and determine what $c$ should be. However, your bounds of integration are very wrong: $y$ cannot be the upper bound of the $y$ variable!

Graph the bounds of integration:

  • the lines $y = c$ and $y=x$
  • the lines $x=y$ (already graphed) and $x=0$
  • shade in the points $(x,y)$ that satisfy the inequalities
  • you should get a region that looks like this ($c=3$)

enter image description here

Hence your bounds should be, if you go for $x$ first, $$ \int_{y=0}^{y=c} \int_{x=0}^{x=y} \frac x y \, \mathrm{d}x \, \mathrm{d}y $$ We choose $\int_{x=0}^{x=y}$ since, if we imagine a horizontal cross-section, then $x$ ranges from $0$ up until the line $x=y$. Within that restriction, such cross-sections exist for each $y$ between $0$ and $c$, justifying the outer integral.

enter image description here

You may wish to review some Calculus III material on multiple integrals if you're struggling with setting up double/triple integrals. Here is a link to Paul's Online Math Notes' material on the matter.

Similarly, for your second function, $$f(x,y)= \begin{cases} \frac{x}{y} & \text{for $0\leq x\leq y\leq 2x$}\\ 0&\text{otherwise}\\ \end{cases}$$ graph:

  • $x=0$ and $x=y$
  • $y=x$ (already done) and $y=2x$
  • shade in the satisfactory points

Then the region of integration looks like so, extending infinitely:

enter image description here

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