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The questions tells us to let X and Y be random variables for which the joint p.d.f. is as follows:

$$f(x,y)= \begin{cases} 2(x+y), & \text{for $0 \le\ y \le\ x \le\ 1$} \\ 0, & \text{otherwise} \\ \end{cases}$$

Find the PDF of $Z=X/Y$

I manage to arrive at the joint pdf:

$$g(z,y)= \begin{cases} 2y^2(z+1), & \text{for ____________} \\ 0, & \text{otherwise} \\ \end{cases}$$

but as you can see, I am unsure of what the restriction should be

furthermore, to get to the marginal pdf of $Z$

$$ \int_{l}^{u} 2y^2(z+1) dy $$

Again, I am unsure of what the limits on the integral ($u$ & $l$) should be. Am I on the right track at all? and how should I then approach the limits/restrictions for this question?

EDIT: I know I was asked to "find the PDF of $Z$", but from what I was taught, I thought we had to first find the joint PDF and then integrate out the other random variables (in this case the $Y$) if I want to find the [marginal] pdf (for $Z$). If this is wrong, can someone produce a worked solution for this then?

EDIT2: Z="X/Y", not "X|Y"

EDIT3: yes, x over y. not x condition y

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    $\begingroup$ You aren't asked to find a joint PDF -- you are asked to find the PDF of $Z$. $\endgroup$ Apr 7, 2014 at 16:15
  • $\begingroup$ How do you define "Z=X|Y"? I have never seen this notation before (except in some MSE question whose author seemed a bit lost). Is this related to conditional probabilities P(A|B) in some sense? $\endgroup$
    – Did
    Apr 7, 2014 at 21:38
  • $\begingroup$ Ach so... this would be a typo: X over Y, not X conditionally on Y. Can the OP confirm this? $\endgroup$
    – Did
    Apr 7, 2014 at 21:48
  • $\begingroup$ Yeah, that typo would be my fault. When I was latex'ing everything, I accidentally messed that up. Sorry! $\endgroup$
    – Hayden
    Apr 7, 2014 at 21:56

2 Answers 2

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First find the distribution of $Z$:

$F_z(z) = P(Z \le z) = P(X / Y \le z) $. Plot in the $X-Y$ plane the aerea represented by $A_z = X/Y \le z$. Then integrate $f(x, y)$ over $A_z$. $$F_Z(z)=P(Z \le z) = \int_{A_z} f(x, y) \ dx \ dy$$

Now you can just differentiate $F_Z$ with respect to $z$ to find the density of $Z$ (if it exists, that is :) )

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Just use the fact that when substituting $x=yz$ then the support will be $0\lt yz\lt y\lt 1$.

$\begin{align}f(x,y) &= 2(x+y)\,\mathbf 1_{0\lt x\lt y\lt 1}\\ g(z,y) &= 2y^2(z+1)\,\mathbf 1_{0\lt yz\lt y\lt 1}\\&= 2y^2(z+1)\,\mathbf 1_{0\lt y\lt 1}\,\mathbf 1_{0\lt z\lt 1}\\h(z) &=\tfrac 23(z+1)\,\mathbf 1_{0\lt z\lt 1}\int_0^1 3y^2\,\mathrm d y\\&=\tfrac 23(z+1)\,\mathbf 1_{0\lt z\lt 1}\end{align}$

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