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I am trying to calculate a CDF of a random variable $x$ which has an upper bound $z$, which is itself a random variable with distribution $G(z)$ on some interval $[z1,z2]$.

E.g. $x \sim U[0,z]$ and $z \sim U[1,2]$.

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Welcome to MSE. Is this homework? If so, tag it as such. –  leonbloy Oct 3 '12 at 16:59
    
Hint: When you write x ~ U[0,z] you actually mean that, for a given value of $z$, $x$ is uniform in $[0,z]$. Which is to say that the conditional probability of $x$ is known. –  leonbloy Oct 3 '12 at 17:02
    
$f(x,z)=f(x|z)f(z)$. Integrate out $z$ and get marginal pdf of $x$. Then integrate to get CDF for $x$. –  Patrick Li Oct 3 '12 at 17:33
    
What this means, if it is to make any sense, is that the CONDITIONAL distibution of $x$ given $z$ is uniform on $[0,z]$. It's easy to neglect to say that, but it's worthwhile to be precise about it. People also often forget to mention independence where it's assumed. –  Michael Hardy Oct 3 '12 at 18:12
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@TAK Yes, you are right that $f(x,z)=\frac{1}{z}$ but be careful about their supports. Since $0<x<z$ and $1<z<2$, the actual joint density is $f(x, z)=\frac{1}{z}\mathbb{I}[0<x<z]\mathbb{I}[1<z<2]$. –  Patrick Li Oct 3 '12 at 19:47
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The conditional pdf of $X$ given $Z$ is $f(x|z)=\frac{1}{z}\mathbb{I}[0<x<z]$. The marginal pdf of $Z$ is $f(z)=\mathbb{I}[1<z<2]$. So the joint pdf of $X$ and $Z$ is $$f(x,z)=f(x|z)f(z)=\frac{1}{z}\mathbb{I}[0<x<z]\mathbb{I}[1<z<2]$$ Then we can integrate over $z$ and get $$f(x)=\int_{\max(x,1)}^{2} f(x,z)dz=\int_{\max(x,1)}^{2}\frac{dz}{z}=\log2-\log(\max(x,1))$$

You can proceed by integrating it to get the CDF of $X$. You need to consider two different conditions when $x > 1$ and $x < 1$.

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Patrick, fantastic, thank you so much, you helped me a lot :):):) –  user43531 Oct 3 '12 at 21:40
    
@TAK Glad to help. Could you accept my answer if you think it's helpful? –  Patrick Li Oct 5 '12 at 12:09
    
Patrick, sorry for the delay. I would be more than happy to accept your answer. But I am new, could you help me to find a way to do so? Do I understand it correctly I need to earn more reputation points first? Many thanks in advance –  TAK Oct 16 '12 at 13:55
    
@TAK The accepting answer option should be somewhere on this page. If you can't find it, that's fine with me. Glad to help. I learn something from this as well. –  Patrick Li Oct 16 '12 at 19:18
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