# Why am I getting the wrong marginal distribution for the minimum of 3 uniforms?

Let $U_1$, $U_2$, and $U_3$ be identical, independent, random variables distributed $Uniform(0, 1)$. Let $L$ be the minimum and $M$ be the maximum of $U_1$, $U_2$, and $U_3$.

I want to find the marginal probability density function (PDF) of $L$ ($f_L(l)$). I got that the joint PDF of $L$ and $M$ is $f_{L, M}(l, m) = 6(m - l)$. I'm pretty sure this is correct.

Next, I tried integrating out the $m$ to get $f_L(l)$:

$$f_L(l) = \int_0^1 6m - 6l~dm \\ = 3m^2 - 6lm \text{ at m = 1 and m = 0.} \\ = 3 - 6l$$

However, I am pretty sure that $f_L(l) = 3(1-l)^2$ just reasoning out what the CDF of $L$ should be $(1 - l)^3$.

Why is my integration wrong?

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The minimum is $\le l$ unless all the $U_i$ are $\gt l$. Thus $$F_L(l)=1-(1-l)^3.$$ That I think is the simplest way to find the cdf of $L$, and hence the density.
But this does not answer your specific question about the integration. The problem there is that the joint density you mention is not quite right. If $l \gt m$, we cannot have density $6m-6l$, the joint density is $0$.
When you are "integrating out" $m$, the variable $m$ should have gone from $l$ to $1$, not from $0$ to $1$. If we calculate $$\int_l^1 (6m-6l)\,dm,$$ we get $(3-6l)-(3l^2-6l^2)$, which is $3(1-l)^2$.