probability density function, integral 1

Question

Let (M,W) be a continous probability space (take M=(a,b)), and let $x:M \rightarrow \mathbb{R} $ be a random variable which is bounded. Prove that

$ \inf_{t\in M} x(t) \le E(x) \le \sup_{t \in M} x(t)$

Hello, We only get a quick introduction on elements of probabilities for our modelling course.We say that W have two properties: $-)W(M)=1$, $-)$if $A \cap B = \emptyset \Rightarrow W(A \cup B)=W(A)+W(B) $. As a example of continous probability space we consider the case for $M=(a,b)$. In this case the probability density function (when it exists) is definied as $f(t):= \lim_{\Delta t \rightarrow 0} \frac{W(t, t + \Delta t))}{\Delta t} $ and for a random real variable $X:M \rightarrow \mathbb{R}, x=x(t)$ we define the expected value of x when the density function exists as $E(x)=\int_a^b x(t) f(t) dt $

Back to the exercise: $\exists K \ge 0: -K \le x(t) \le K \forall t \in M$

It follows that $-K \le \inf_{t \in M} x(t) \le x(t) \le \sup_{t\in M} x(t) \le K \forall t \in M $

So $\inf_{t \in M} x(t) \int_a^b\lim_{\Delta t \rightarrow 0} \frac{W(t, t + \Delta t))}{\Delta t}dt \le E(X) \le \sup_{t\in M} x(t)\int_a^b\lim_{\Delta t \rightarrow 0} \frac{W(t, t + \Delta t))}{\Delta t}dt$

We don´t discuss that the integral of the density function should be 1 on the whole intervall, we only defined it as above.

Now i must prove: $\int_a^b\lim_{\Delta t \rightarrow 0} \frac{W(t, t + \Delta t))}{\Delta t}dt=1$

For this i want to use the fundamental theorem of calculus. $ \lim_{\Delta t \rightarrow 0} \frac{W((t,t+\Delta t))}{\Delta t}= \lim_{\Delta t \rightarrow 0} \frac{1- W((a,t] \cup [t+ \Delta t,b))}{\Delta t}= \lim_{\Delta t \rightarrow 0} \frac{1- W((a,t]) - W( [t+ \Delta t,b))}{\Delta t}=\lim_{\Delta t \rightarrow 0} \frac{-W((a,t]) + W((a,t + \Delta t))}{\Delta t}$

Now i define $p:t \rightarrow W((a,t))$

fundamental theorem of calculus: $ \int_a^b p'(t) dt = p(b)-p(a)=W((a,b))-W((a,a))=1- W((a,a))$

But my problem is that i must show that the probability of one discrete point is $0$. How i can do these?

Answer 1

By definition, probability measure $W$ has density $f: [a,b] \to \mathbb R$ if and only if $$W(a,x)=\int_a^x f(t) dt,\quad x \in (a,b)$$

Your course may be using slightly different (but equivalent) definition of the density of probability measure; your question does not contain this definition — see more about this at the end of this answer.

If probability measure $W$ does have density $f$, then from this definition, it immediately follows that $$\int_a^b f(t) dt=W(a,b)=1$$

Let us prove, nevertheless, that $W(\{t\})=0$ for every $t \in (a,b)$ whenever probability measure $W$ has density $f$. Well, then $$W(\{t\})=\int_a^b \delta_t(x) f(t) dt=0$$ where $\delta_t(x)$ is $1$ when $x=t$ and zero otherwise. Why did I mention previosly that function $$F(x)=W(a,x)=\int_a^x f(t) dt$$ is continuous on $(a,b)$? I was thinking about the following proof that uses $\sigma$-additivity of probability measures.

You said that all you know about probability measure $W$ is that

1. $W(a,b)=1$
2. $A \cap B=\emptyset \implies W(A \cup B)=W(A)+W(B)$

Actually, the properties 1-2 are not enough for $W$ to be a probability measure — you need something more, a countable additivity (also called $\sigma$-additivity):

3. $A_i \cap A_j = \emptyset, \ i \neq j \implies W ( \bigcup_{i = 1}^{\infty} A_i) = \sum_{i = 1}^{\infty} W (A_i)$

for any sequence $A_1,A_2,\ldots$ of measurable sets. I will use this property of probability measure $W$ to prove that $W(\{t\})=0$. Indeed, if $t_1,t_2,\ldots \in (a,b)$ is sequence such that $t_i \to 0$, we can write the singleton set $\{t\}$ as $$ \{t\}= \bigcap_{i = 1}^{\infty} [t, t + t_i) = [t, b) \backslash \bigcup_{i = 1}^{\infty} [t + t_i, b) = [t, t + t_1) \backslash \bigcup_{i = 1}^{\infty} [t + t_{i + 1}, t + t_i) $$

Sets forming last union above are disjoint, and from 3. we have that $$ W\{t\}= W [t, t + t_1) - \sum_{i = 1}^{\infty} W [t + t_{i + 1}, t + t_i) = 0 $$ since $$\begin{aligned} \sum_{i = 1}^k W [t + t_{i + 1}, t + t_i) &= W [t + t_{k + 1}, t + t_1) \\ &= F (t + t_1) - F (t + t_{k + 1}) \to W [t, t + t_1) \end{aligned}$$

Let us now return to the definition of density. If function $F(x)$ is differentiable on $(a,b)$, then probability measure $W$, by fundamental theorem of calculus, has density $$f(t)=F'(t)=\lim_{\Delta t \to 0} \frac{F(t+\Delta t)-F(t)}{\Delta t} =\lim_{\Delta t \to 0} \frac{W[t,t+\Delta t)}{\Delta t} =\lim_{\Delta t \to 0} \frac{W(t,t+\Delta t)}{\Delta t}$$ The last equation hold, because we now know that $W(\{t\})=0$. From the question, I got the impression you were using the last limit as the definition of density, but the existence of this limit for every $t \in (a,b)$ does not necessarily mean that measure $W$ has density. For example, if we define $W$ to be Dirac measure $$W(A)=\begin{cases} 1, & \text{if}\ \frac{1}{2} \in A \\ 0, & \text{otherwise} \end{cases}$$ then $W(t,t+\Delta t)=0$ for every $t \in (a,b)$ and sufficiently small $\Delta t$ and the limit above exists for every $t \in (a,b)$ and is equal to zero, but this measure does not have density.