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Given the variables $X$ and $Y$ which are correlated and jointly Bivariate with corresponding probability distribution function denoted by $f_{X,Y}(x,y)$, and given the linear relationship

$Z = a+\frac{b}{c}Y,\quad a,b,c\in\mathbb{R}$

How do I get the joint probability distribution $f_{X,Z}(x,z)$.

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    $$P\{X \leq x, Z \leq z\} = P\{X \leq x, a + (b/c)Y \leq z\} = P\{X \leq x, Y \leq c(z-a)/b\}$$ assuming $(b/c) > 0$.2012-02-12
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    @Dilip: Thanks, that means $P\{X=x,Z=z\}=\frac{c}{b}P\{X=x,Y=\frac{c}{b}(z-a)\}$...2012-02-12
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    I don't know how you got the result $$P\{X=x,Z=z\}=\frac{c}{b}P\{X=x,Y=\frac{c}{b}(z-a)\}$$ and I am not at all sure that it follows from what I wrote. Be careful! If $c/b$ is large enough, you might well get a probability larger than $1$.2012-02-12
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    I got the result by taking the derivative with respect to z, i.e, to obtain the probability density function $ f_{X,Z}(x,z)$ from the cumulative distribution function $ F_{X,Z}(x,z)$ , that is to say $f_{X,Z}(x,z)=\frac{\partial^2 }{\partial x\partial z}F_{X,Z}(x,z)$2012-02-13
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    >I got the result by taking the derivative.... BUT that is _not_ what you wrote; you obtained $f_{X,Z}(x,z)$ but that **is not the same** as $P\{X=x,Z=z\}$ which denotes the probability that $X$ has value $x$ and simultaneously $Z$ has value $z$. For continuous random variables, $P\{X=x,Z=z\} = 0$ for all $x$ and $z$.2012-02-13

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One gets $$ f_{X,Z}(x,z)=|c/b|\cdot f_{X,Y}(x,(c/b)(z-a)). $$ More generally, if $(T,Z)=M\cdot(X,Y)$ and $M$ is invertible, $$ f_{T,Z}(t,z)=|\det M|^{-1}\cdot f_{X,Y}(M^{-1}\cdot(t,z)). $$

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    Thanks @Dider, so that means what I wrote is right2012-02-16