Is my distribution function strictly increasing?

Question

I'm trying to figure out if a function is strictly increasing.

I begin by introducing the information I know:

We are given a non-negative random variable X with strictly increasing cumulative distribution function $F_X$, i.e. $P(e<X\leq f)>0$ for all $0\leq e<f$.

Show that $F_{I(X)}$ is strictly increasing, i.e. $P(g<I(X)\leq h)>0$ for all $0\leq g<h$.

where

\begin{equation} I(X)= \begin{cases} X & \text{for } X< a, \\ a & \text{for } a\leq X\leq b, \\ X-(b-a) & \text{for }X>b. \end{cases} \end{equation} and $0<a<b<\infty$.

$\textbf{My attempt}$:

First we note that we have 3 scenarios to consider: \begin{equation} P(g<I(X)\leq h)= \begin{cases} P(g<X\leq h) & \text{for } X< a, \\ P(g<a\leq h) & \text{for } a\leq X\leq b, \\ P(g<X-(b-a)\leq h) & \text{for }X>b. \end{cases} \end{equation}

$\textbf{1:}$ If $X<a$ then let $h=a$, then by assumption $P(g<X\leq a)>0$ for all $0\leq g<a$.

$\textbf{2:}$ If $X>b$ then we have by assumption that $P(g<X-(b-a)\leq h)=P(g+(b-a)<X\leq h+(b-a))>$ for all $0\leq g+(b-a)<h+(b-a)$.

$\textbf{3:}$ If $a\leq X\leq b$ then there exists an $a$ such that $P(g<a\leq h)=0$, to see why pick $a=50$, $g=0$ and $h=40$ then $a\notin (g,h]$ with probability 1.

I'm not really sure what to conclude with, I'm not even sure if it makes any sense to compute $P(g<a\leq h)$, as this isn't really a stochastic event

Answer 1

Write $Y:=I(X)$. First gain some intuition about what $Y$ is doing. The difference between $Y$ and $X$ is that $Y$ "holds onto" the value $a$ when $a\le X\le b$, then "continues where $X$ left off" as soon as $X>b$.

You want to show that $P(g<Y\le h)>0$ for all $0\le g<h$. It's right for you to break into three scenarios, but in cases 1 and 3 you are not allowed to force $h$ or $a$ to take any specific value.

Instead, it is better to break into scenarios according to where $a$ lives in relation to the interval $(g,h)$:

1. If $a$ is to the right of $(g,h)$ (i.e., $h\le a$), then the event $\{g < Y \le h\}$ is the same as the event $\{g < X \le h\}$, which has positive probability.

2. If $a$ is to the left of $(g,h)$ (i.e., $a\le g$), then the event $\{g < Y\le h\}$ is the same as the event $\{g<X-(b-a)\le h\}$. Your job is to show that this last event has positive probability.

3. If $a$ is inside the interval $(g,h)$, the argument is a bit more intricate: Observe that the event $\{g < Y \le h\}$ contains the event $\{Y=a\}$, which equals the event $\{a\le X\le b\}$, which has positive probability.

Answer 2

I think you should reason about I(X) rather than about X.

Given e and f with $0 \le e \lt f$

if $a \le e$ or $a \gt f$ then yout can easily find $0\le e_1 \lt f_1$ where

$e \lt I(X) \le f \Leftrightarrow e_1 \lt X \le f_1$

so $P(e \lt I(X) \le f) = P(e_1 \lt X \le f_1) \gt 0$

otherwise ( $ e \lt a \le f$) : $P(e \lt I(X) \lt f) \ge P(I(X)=a) = P(a\le X \le b) \ge P(a\lt X \le b) > 0$

Answer 3

Your attempt isn't quite right.

The cases $X<a, a\le X\le b, X>b$ in the definition of $I(X)$ make sense because $I(X)$ is a random variable whose value depends on the value of the random variable $X$. (It can be confusing that the symbol $X$ is used for specific values as well as to refer to the random variable as a whole.)

In contrast, $P(g<I(X)\le h)$ is not a random variable. It's a constant quantity that takes into account the whole distribution of $X$, so we can't have cases for $P(g<I(X)\le h)$ that depend on $X$. Instead, you'll want cases comparing the constants $g,h,a,b$ to each other.

For example, if $a\le g$, then we're looking at an interval of $I(X)$ values that are all larger than $a$, and these values come from $X$ values larger than $b$. Specifically, whenever $g+b-a<X\le h+b-a$, we have $g<I(X)\le h$, so $P(g<I(X)\le h) \ge P(g+b-a<X\le h+b-a)$, and this is a nonempty interval of $X$ values so the probability is nonzero.

Answer 4

It is not clear what the meanings are of your cases for $P(g<I(X)\leq h)$. The idea of that particular notation of probability is that $X$ could be any value in its range and the probability $P(g<I(X)\leq h)$ should be defined even though we have no other information about the value of $X.$ So it seems a very strange idea that we have to know whether $X<a$ before we can say what the probability is.

What I think you might have had in mind is actually $$P(g<I(X)\leq h \mid X < a)$$ for the first case, that is, a conditional probability, which allows you to assume that $X<a$ when working out that particular probability. Likewise the other two cases can be written as conditional probabilities. Then you can use the fact that the conditions are a disjoint partition of the possible values of $X$, and what you know of probabilities such as $P(X<a)$, to assemble these probabilities into the desired total probability $P(g<I(X)\leq h)$.

Another way to look at this is to think about how a graph of $F_{I(X)}$ would look. To the left of $a$ it is the same as $F_X,$ but at $a$ it suddenly jumps upward, and the rest of it to the right of $a$ is like the graph of $F_X$ to the right of $b$, but shifted leftward to start at $a$ instead of $b$.

Looking at this graph, it should be clear how to conclude that $P(g<I(X)\leq h)$ when $a$ is outside $(g,h]$ on the right-hand side, and just a little more work is required when $a$ is completely outside $(g,h]$ on the left side. The case where $a$ is in the interval $(g,h]$ actually is the easiest of all, provided that you remember you just need to show it is positive, not to exhibit exactly how to find its value.