# Does taking minimum preserve density monotonicity?

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.

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:

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 $$g$$:

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 $$(0,1)$$.

• 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 $t\in(2/3, 1)$ Apr 5, 2022 at 6:22
• 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. Apr 5, 2022 at 6:36
• @Nikolay : Sorry for the mistake. Now I have another counterexample. Apr 5, 2022 at 7:25
• Thanks for this great counterexample! Apr 5, 2022 at 18:29