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Suppose $X$ and $Y$ are continuous random variables with a joint density function $f_{X,Y}$. Both $X$ and $Y$ are supported on $(0,1)$ and have continuous (can be assumed differentiable) and non-increasing densities $f_X$ and $f_Y$ respectively. $f_X(x)>0$ and $f_Y(x)>0$ for $x\in(0,1)$. Let $M = \min\{X,Y\}$. Is $f_M(t)$, the density of $M$, necessarily non-increasing?

The conclusion is straightforward when $X$ and $Y$ are independent but I do not see how to prove it in the presence of unknown dependence. I could not find a counterexample as well.

Here are a couple of approaches that I tried.

  1. Clearly, the CDF of $M$ can be expressed as $$F_{M}(t) := \Pr(M\le t)=1 - \int_{t}^1\int_{t}^1f_{X,Y}(x,y)dxdy.$$ Therefore, $$f_M(t) = \int_t^1f_{X,Y}(t,y)dy+\int_t^1f_{X,Y}(x,t)dx$$ and the derivative is $$f'_M(t) = \int_t^1\frac{\partial f_{X,Y}(t,y)}{\partial t}dy+\int_t^1\frac{\partial f_{X,Y}(x,t)}{\partial t}dx - 2f_{X,Y}(t,t).$$ I do not see how one can use the monotonicity conditions on the marginal densities to make progress from here.

  2. Alternatively, we could write $\Pr(M\le t)$ as $$\Pr(M\le t) = \Pr(X\le t) +\Pr(Y\le t) - \Pr(\max\{X,Y\}\le t)$$ and consequently $$f_M(t) = f_X(t)+f_Y(t) - \frac{d}{dt}F_{X,Y}(t,t),$$ where $F_{X,Y}$ is the joint CDF. As far as I understand, for any $X$ and $Y$ we have a bivariate copula function $C(\cdot,\cdot)$ such that $F_{X,Y}(t,t) = C(F_X(t), F_Y(t))$ and therefore $$f'_M(t) = f'_X(t)(1-C_1)+f'_Y(t)(1-C_2)-(f^2_X(t)C_{11}+2f_X(t)f_Y(t)C_{12}+f^2_Y(t)C_{22}),$$ where $C_1$ is the derivative of $C(\cdot, \cdot)$ with respect to the first argument evaluated at $(F_X(t), F_Y(t))$ and $C_2, C_{11}, C_{12},$ and $C_{22}$ are defined analogously. The first two terms in the equation above are negative but it is not clear how to deal with the rest. I could not find any useful results in the copulas literature.

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1 Answer 1

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The answer is no. Indeed, let $X\sim U(0,1)$, where $U(0,1)$ is the uniform distribution on $[0,1]$, and let $Z$ be an independent copy of $X$. Next, let $$Y:=\frac Z2\,1(X<1/2)+\frac{1+Z}2\,1(X>1/2).$$ So, the distribution of the random point $(X,Y)$ is the half-and-half mixture of the uniform distribution on the square $(0,1/2)^2$ and the uniform distribution on the square $(1/2,1)^2$. That is, the distribution of the random point $(X,Y)$ is the uniform distribution on the union of the "gray" squares $(0,1/2)^2$ and $(1/2,1)^2$, shown here:

enter image description here

Then $Y\sim U(0,1)$. So, the common density of $X$ and $Y$ is nonincreasing on $[0,1]$.

One the other hand, $M=\min(X,Y)$ has the non-monotonic density $g$ on $[0,1]$ such that $g(m)=2-4m$ for $m\in(0,1/2)$ and $g(m)=4-4m$ for $m\in(1/2,1)$.

Here is the graph of $g$:

enter image description here


The idea of this example is that here, as can be seen from the two-gray-squares picture, the density $g$ of $M$ is close to $0$ in a left neighborhood of $1/2$, which makes the monotonicity of $g$ impossible.


In this example, the pdf of $M$ is bimodal and has one (near-)zero in $(0,1)$. Similarly, for any $n\ge2$, by having $n$ non-overlapping "gray" squares cut in half by the diagonal of the unit square, we can make a saw-like pdf of $M$ with $n$ modes and $n-1$ (near-)zeroes in $(0,1)$.

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  • $\begingroup$ Thanks a lot for your answer! I agree that $X$ and $Y$ are uniformly distributed on [0,1] but I think that there is a mistake in the density of the minimum. For example, consider $\Pr(\min(X,Y)\le 1/3)$. This probability is the probability of 3 dark-brown squares plus the probability of 2 light-brown squares. This gives us 7/12. On the other hand, from your graph this probability is 1/6. In general, I calculated for your example that $\Pr(\min\{X,Y\}\le t)$ equals $2t-3t^2/4$ for $t\in (0, 1/3]$, $1/4+t$ for $t\in(1/3, 2/3]$ and $1/4+3t/2 - 3t^2/4$ for $t\in(2/3, 1)$ $\endgroup$
    – Nikolay
    Apr 5, 2022 at 6:22
  • $\begingroup$ The implied density function is $(2-3t/2)1\{0<t\le 1/3\}+$ $1\{1/3<t\le 2/3\}+$ $3/2(1-t)1\{2/3<t<1\}$ which is non-increasing. $\endgroup$
    – Nikolay
    Apr 5, 2022 at 6:36
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    $\begingroup$ @Nikolay : Sorry for the mistake. Now I have another counterexample. $\endgroup$ Apr 5, 2022 at 7:25
  • $\begingroup$ Thanks for this great counterexample! $\endgroup$
    – Nikolay
    Apr 5, 2022 at 18:29

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