Showing that the difference between two CDF functions is positive when CDFs are symbolic expressions

I'd like to check the sign of a symbolic expression conditional on some predefined assumptions. More specifically, first, I want to assume that the mean and variance of a normally distributed random variable is between 0 and 1. Then I want to show that the probability that this variable is between two values (that are also between 0 and 1) is positive. Here is my code:

$Assumptions = {1>kmax>kmin>0,1>m>0,1>sigma>0}; FullSimplify@Positive[CDF[NormalDistribution[m,sigma],kmax]-CDF[NormalDistribution[m,sigma],kmin]]  which returns Can someone help me understand why Mathematica does not see that this expression is strictly positive? Thanks in advance! • I think you are going about this the wrong way. The question is not really a Mathematica question, nor a Normal distribution question. If$X$is a continuous random variable with cdf$F(x)$, then by definition$F(x)$is a non-decreasing function of$x$, and strictly increasing over any positive density. Finally,$P(a<X<b) = F(b)-F(a)$for$b>a\$... and that covers you for everything. Commented Sep 28, 2023 at 14:10
• Thank you. I am aware of the mathematical proof you provide. However, my end goal is to use Mathematica to conveniently check the signs of other more complex symbolic expressions I have at hand. The fact that Mathematica fails to recognize the positivity of the simple expression above means that it can't help me assess other expressions either. Commented Sep 29, 2023 at 10:48

The documentation clearly says that Positive works only for numbers. This explains the output obtained by you. That you want can be done as follows.
FunctionMonotonicity[CDF[NormalDistribution[m, sigma], x], x,StrictInequalities -> True]

1
• @elifcansu: I don't know such a function in Mathematica. FindInstance may do the job in some cases That's all. Commented Sep 27, 2023 at 17:08