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What could be a general way to find the Joint PDF given two PDFs?

For example, $X$ and $Y$ be the two random variables with PDFs:

$f(x)$ = $\{$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ {1\over 40}$; if $0 < x < 10$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ $$0$ ;if $10 < x < 30$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ $$1\over 40$ ;if $30 < x < 60$

$\ \ \ \ \ \ \ \ \ \ \ \ \}$

$f(y)$ = $\{$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ $$0$; if $0 < y < 10$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ $$1\over 20$ ; if $10 < y < 30$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ $$0$; if $30 < y < 60$

$\ \ \ \ \ \ \ \ \ \ \ \ \ $$\}$

What is the way to find $f(z)$ $?$, where $Z$ is a continuous random variable made* up of $X$ and $Y$ $?$

*made up: if I am making any sense

user118494
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Bibrak
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    Without further assumptions one cannot do it. But if $X$ and $Y$ are independent, you multiply. – André Nicolas Sep 23 '15 at 04:55
  • @AndréNicolas Yes they are independent. I am confused how to multiply. Is it what f(x,y) = xy/800 ?? – Bibrak Sep 23 '15 at 04:56
  • For example, the joint density function $f(x,y)$ is $(1/40)(1/20)$ if $x$ is between $0$ and $10$ and $y$ is between $10$ and $30$. There are a number of other cases. For instance it is $(1/40)(0)$ if $x$ and $y$ are both between $0$ and $10$, and there are other "rectangles" on which it is $0$. – André Nicolas Sep 23 '15 at 05:00
  • @AndréNicolas and then how to find expected value from such a pdf? – Bibrak Sep 23 '15 at 09:20
  • You asked for the joint pdf. What is the random variable you want to compute the expectation of? Presumably some function of $X$ and $Y$, like $W=\sqrt{X^2+Y^2}$, Then $E(W)=\iint \sqrt{x^2+y^2}f(x,y),dx,dy$. – André Nicolas Sep 23 '15 at 13:24
  • If you wanted the distribution of some random variable $W=g(X,Y)$ like the one above, you can use the joint density $f(x,y)$ to find the cdf of $W$, and then differentiate. Or else in some cases you can use convolutions. – André Nicolas Sep 23 '15 at 13:29

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