3
$\begingroup$

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 $z \sim U[1,2]$.

  • 0
    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.2012-10-03
  • 0
    $f(x,z)=f(x|z)f(z)$. Integrate out $z$ and get marginal pdf of $x$. Then integrate to get CDF for $x$.2012-10-03
  • 0
    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.2012-10-03
  • 0
    Guys, thank you so much. I tried that way, but seem to be making a mistake. In my example, f(x|z)=1/z and f(z)=1, right? Then f(x,z)=1/z and thus f(x)=ln2. But since the support for x is [0,2], the resulting F(x) is not a distribution. Moreover, logically (and using simulations), F(x) in my example should be a positively skewed function (not linear). What am I doing wrong?2012-10-03
  • 1
    @TAK Yes, you are right that $f(x,z)=\frac{1}{z}$ but be careful about their supports. Since $0 and $1, the actual joint density is $f(x, z)=\frac{1}{z}\mathbb{I}[0.2012-10-03

1 Answers 1

1

The conditional pdf of $X$ given $Z$ is $f(x|z)=\frac{1}{z}\mathbb{I}[0. The marginal pdf of $Z$ is $f(z)=\mathbb{I}[1. So the joint pdf of $X$ and $Z$ is $$f(x,z)=f(x|z)f(z)=\frac{1}{z}\mathbb{I}[0 Then we can integrate over $z$ and get $$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 $x < 1$.

  • 0
    Patrick, fantastic, thank you so much, you helped me a lot :):):)2012-10-03
  • 0
    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 advance2012-10-16
  • 0
    @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.2012-10-16