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I know that to find the marginal density I need to compute

$f(x) = \int_{-\infty}^{\infty}f(x,y)dy$

But when I have i.e.

$f(x,y) = (1/θ^2)e^{-y/θ}$ for $0<x<y<\infty$

Then to my understanding it is calculated

$f(x) = \int_{0}^{x}(1/θ^2)e^{-y/θ}dy$

$f(y) = \int_{0}^{y}(1/θ^2)e^{-y/θ}dx$

Yet on some resources I find the integral going from i.e. $\int_{x}^{\infty}$

What is the general rule for the boundaries?

ReRed
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  • Okay yeah you are right, it was a stupid example function. But the main question was how I define the boundaries. I'll update the question. – ReRed Aug 21 '19 at 08:58

1 Answers1

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If $f(x,y)$ is density function concentrated on $x>0,y>0$ then the first marginal is $f(x)=\int_0^{\infty} f(x,y)dy$. In case $f(x,y)=0$ for $y <x$ this reduces to $f(x)=\int_x^{\infty} f(x,y)dy$ because the integral from $0%=$ to $x$ vanishes.

Kavi Rama Murthy
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  • Hmm, okay. The updated question is exactly the solution from my textbook except for the f(x), as only f(y) was asked. For f(y) the textbook says: $f(y) = \int_{0}^{y}(1/θ^2)e^{y/θ}dx$, but according what I understand from your point, it should be $f(y) = \int_{y}^{\infty}(1/θ^2)e^{y/θ}dx$, no? – ReRed Aug 21 '19 at 09:09
  • Sorry I had another edit my main question and the comment above this one. – ReRed Aug 21 '19 at 09:11
  • There are two marginals and you should not use the same symbol $f$ for both. Secondly your integrals do not make sense. If you are integrating w.r.t. $y$ the limits if integration should not depend on $y$. – Kavi Rama Murthy Aug 21 '19 at 09:13
  • You mean this expression $f(y) = \int_{0}^{y}(1/θ^2)e^{y/θ}dx$ doesn't make sense? It is like that according to my textbook and professor... – ReRed Aug 21 '19 at 09:16
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    In the first integral the limits of integration have to be determined from the given condition $0<x<y<\infty$. Thus, for any $x>0$, $y$ ranges from $x$ to $\infty$ so you have to integrate from $x$ to $\infty$ not from $0$ to $x$. – Kavi Rama Murthy Aug 21 '19 at 09:27
  • Okay, so $f_x(x)$ with the condition $0<x<y<\infty$ we have $x>0$, therefore $y$ ranges from $x$ to $\infty$, but for $f_y(y)$, we have $y<\infty$, therefore we integrate from $0$ to $y$. Did I understand it correctly? – ReRed Aug 21 '19 at 09:32
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    Yes, you got the point. – Kavi Rama Murthy Aug 21 '19 at 09:35