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I have a question. Let's suppose that the two random variables $X1$ and $X2$ follow two Uniform distributions that are independent but have different parameters:

$X1 \sim Uniform(l1, u1)$

$X2 \sim Uniform(l2,u2)$

If we define X3 as the maximum of X1, and X2, i.e., $X3 = max(X1, X2)$, what kind of distribution would it be? It is certainly not a uniform and I can calculate the cumulative probability, i.e., $p(X3 <= a)$, because $p(X3 <= a) = p(X1 <= a) p(X2 <= a)$. But, I wonder if there exists any density function that can model X3.

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1 Answer 1

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The distribution of $Z = \max(X,Y)$ of independent random variables is $$F_Z(z) = P\{\max(X,Y)\leq z\} = P\{X \leq z, Y \leq z\} = P\{X\leq z\}P\{Y \leq z\} = F_X(z)F_y(z)$$ and so the density is $$f_Z(z) = \frac{d}{dz}F_Z(z) = f_X(z)F_Y(z) + F_X(z)f_Y(z).$$

All of this holds regardless of the distributions of $X$ and $Y$, that is, they need not be uniformly distributed random variables. But, for uniform distributions, the density of $Z$ has simple form since $f_X(z)$ and $f_Y(z)$ are constants and $F_X(z)$ and $F_Y(z)$ are constants or linearly increasing functions of $z$.
For example, if $X, Y \sim U(0,1)$, then $f_Z(z) = \begin{cases}2z, &0 < z < 1,\\ 0, &\text{otherwise},\end{cases}$ which is not a uniform distribution. On the other hand, if $X\sim U(a,b)$ and $Y\sim U(c,d)$ where $c >b$ and so the value of $Y$ is always larger than the value of $X$, then $Z = max(X,Y) = Y$ is uniformly distributed on $(c,d)$. That is, your notion that $\max(X,Y)$ "is certainly not uniform" should be tempered with a little less certainty.

Exercise: work out the density for the cases when $(a,b)$ and $(c,d)$ overlap partially, and when one is a subinterval of the other.

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