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I am trying to solve this question that asks for a new variable V such that V = $\frac{Y}{X}$. The given information is $\begin{align*} f(x,y)&=\frac{2x+y}{36} & 0 \leq y \leq x, \hspace{5mm} x + 2y \leq 6 \\ \end{align*}$ . My Issue is I can't figure out the bounds/cases for the new variable V.

The first bound is easy. From $0 \leq y \leq x$ it become $0 \leq y/x \leq 1$. After this I cannot figure it out.

StubbornAtom
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air bmx
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1 Answers1

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You want to know the probability distribution of $V$. Let's try to find the cumulative distribution function:

$$\mathbb{P}(V\leq v) = \mathbb{P}(Y\leq vX) = \int_{y \leq vx}\tilde{f}(x,y)dxdy$$

With

$$\tilde{f}(x,y) = \begin{cases}f(x,y) \; , & \text{ if } \; 0 \leq y \leq x, \hspace{1mm} x + 2y \leq 6 \; , \\ 0 \; , & \text{ else} \; .\end{cases}$$

I've made a little plot to visualize the integration domain

Picture of integration domain

I hope it's clear enough, but you can see that the range of $V$ is $[0,1]$. To do the computation, you will need to find the intersection between the lines represented by $y = vx$ and $x+2y=6$. Let's call it $(x_v,y_v)$. Then your integral can be reformulated as

$$\mathbb{P}(V\leq v) = \int_{x=0}^{x_v}\int_{y=0}^{vx}\frac{2x+y}{36}dydx + \int_{x=x_v}^{6}\int_{y=0}^{3-x/2}\frac{2x+y}{36}dydx$$

Can you take it from here?

Raskolnikov
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  • Yes I can thank you very much. Just to confirm, once I find the CDF, I can find the PDF by taking the derivative correct? – air bmx Feb 19 '20 at 04:24
  • Indeed. If you are familiar with taking derivatives of integral forms, you can even try that before trying to work out the integrals. – Raskolnikov Feb 19 '20 at 05:32