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I have the following distribution:

$$P(Z \leq z ) = \begin{cases} z/20 & 0 < x <10 \\ 1/2 + z/200 & 10 \leq z < 110 \\ 1 & z > 110 \end{cases} $$

How can I write this as a mix of two uniform distributions? For reference, the cdf of U(a,b) is (x - a)/(b - a), and the pdf is 1/(b - a).

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    Should $z$ be $x$? And this function can't be a cumulative distribution function since it's not increasing. Perhaps you mean for it to be a density function? – Greg Martin Sep 18 '19 at 05:42
  • fixed @GregMartin –  Sep 18 '19 at 09:26
  • I think maybe it's still wrong. I'm trying to find the equilibrium distribution of a random variable that is $10$ with probability $0.1$ and $110$ with probability $0.1$. Can you help me? That is what the distribution function is supposed to be –  Sep 18 '19 at 09:37
  • The equilibrium distribution is $X$ is $F_{e}(x) = \frac{1}{E(X)} \int_{0}^x P(X > x)$. –  Sep 18 '19 at 09:37
  • If you mean the underlying variable is 10 with probability 0.9, I think you have the right formula. The probabilities 0.1 and 0.1 don't add up to 1, so that cannot be what you wanted. – David K Sep 18 '19 at 10:55
  • @DavidK I typoed. I meant $X$ takes on $10$ with probability $0.1$ and $110$ with probability $0.9$. I want to find $F_{e}(x)$ with the formula I gave in the other comment –  Sep 18 '19 at 11:50
  • In that case the expected value of $X$ is $100$ and you need a different distribution than shown. – David K Sep 18 '19 at 21:02

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