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Consider a continuously differentiable and strictly decreasing function $G:[0,T]\rightarrow [0,1]$ for some $T>0$ with $G(0)=1,G(T)=0$.

Suppose you normalize $G$ with the area under its graph and define $f(t)$ to be

$$f(t)=\frac{G(t)}{\int^T_0G(x)dx} \text{ if $t\in(0,T)$, 0 otherwise.}$$

My questions:

  1. Why does normalizing G by the entire area under the curve produce something that behaves exactly like a probability density function?

  2. What is the intuition behind #1?

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2 Answers 2

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Theorem. Let $f$ be a non-negative function (Lebesgue) integrable over some nice (Borel) set $\mathrm{V}$ of $\mathbf{R}^d.$ There exists a probability space $(\Omega, \mathscr{F}, \mathbf{P})$ and a random vector $X:\Omega \to \mathbf{R}^d$ such that $\mathbf{P}(X \in \mathrm{A}) = c \int\limits_\mathrm{A} f,$ where the normalising constant is $c^{-1} = \int\limits_\mathrm{V} f.$

Proof. Let $\Omega = \mathrm{V},$ $\mathscr{F}$ the Borel sets of $\mathrm{V}$ and define $\mathbf{P}$ on $\mathscr{F}$ by $\mathbf{P}(\mathrm{A}) = c \int\limits_\mathrm{A} f,$ with $c$ as stated in the theorem. Standard properties of Lebesgue integral show that $\mathbf{P}$ is indeed a probability measure (it is a positive measure with total mass equal to unity). To construct $X$ simply consider $X$ to be the identity function from $\Omega$ into $\mathbf{R}^d.$ Q.E.D.

Comment: There is very little intuition behind this construction except that is a theorical results that says the classical probability/statistics taught in first year college (all by densities) actually is a particular case of the abstract probability theory of graduate school using measure theory as backbone; so what is taught and done is very well justified by measure theory which is really quite flexible. What I think is really quite nice of the explicit construction is that it shows that the concept of random vector (which is primordial in elementary courses) actually is somewhat arbitrary (literally we are using the identity function as random vector). What really is important is that a concept such as a normalised positive measure captures very well the intuitive meaning of chances with the frequency intuition ("we say something has $x\%$ chance of occuring if in the long run of repeated experiments, about $x\%$ of them are success"). Furthermore, it also shows that the same space of measurable sets, namely the pair $(\Omega, \mathscr{F})$, is really meaningless probabilistically until we add a nice probability measure on top of it.

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  • $\begingroup$ William, this helped me a lot. Thanks again! $\endgroup$ Feb 1, 2022 at 19:38
  • $\begingroup$ Do you have a reference of the theorem you mention somewhere like in Billingsley (1976)? $\endgroup$ Feb 1, 2022 at 19:39
  • $\begingroup$ The only book I know that made probability theory very formally is (Robert B.) Ash's book "Real Analysis and Probability". Be warned, Ash works 4 chapter on measure theory and analysis before doing probability theory. The payoff is huge: you get all probability theory from a theoretical perspective (e.g. the theorem I stated and MUCH more). $\endgroup$
    – William M.
    Feb 1, 2022 at 19:41
  • $\begingroup$ I know Ash's book, and it is excellent! I will take a look, but it would be helpful to know from which Chapter this comes from. Thanks. $\endgroup$ Feb 1, 2022 at 19:44
  • $\begingroup$ I just checked my book, Ash makes this a comment on p. 209 (after definition 5.6.2). Be sure to use "Real Analysis and Probability" that Ash wrote like crazy (he wrote many books while alive). $\endgroup$
    – William M.
    Feb 1, 2022 at 19:48
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Any nonnegative function $f$ that integrates to 1 is the pdf of some random variable — namely, the random variable with cdf $F(x) :=\int_{-\infty}^x f(x) dx$.

You simply went straight to defining a density — your normalized $G(x)$ — as opposed to motivating it with some random variable or cdf.

The key step here is that you ensured that the normalized function (let's call it $g(x)$) integrates to 1 over $[0,T]$, hence you can answer questions like $P(X \in [0,0.5T])= \int_0^{0.5T}g(x)dx$

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  • $\begingroup$ Hi Bey, thanks for the response. Yes, I just checked whether the way I defined $f(t)$ in fact satisfies the definition of pdf for a continuous random variable. But what is the intuition behind dividing a strictly decreasing function that is capped between 0 and 1 by $G(x)$ (i.e. the area under that curve) produces a pdf? Any thoughts? $\endgroup$ Feb 1, 2022 at 13:27
  • $\begingroup$ Also, is the random variable here $t\in[0,T]$ or $Q(t)\in[0,1]$? $\endgroup$ Feb 1, 2022 at 13:30
  • $\begingroup$ @FrankSwanton your random variable is $t \in [0,T]$. The fact that its range is $[0,1]$ isn't really important, it could very well be $[100, 7000]$ - it's the normalization that makes it a pdf. You've simply defined a random variable that is more likely to be closer to 0 than T. $\endgroup$
    – Annika
    Feb 1, 2022 at 13:47
  • $\begingroup$ Ok, great. If you could elaborate why the normalization makes it a pdf, just some intuition and edit your response, I will complete selecting your answer. It is just that your response doesn't necessarily shed light on what the intuition behind the normalization makes it a pdf. Thanks! $\endgroup$ Feb 1, 2022 at 15:39
  • $\begingroup$ @FrankSwanton I guess I'm not sure what kind of intuition you are aiming to get. By definition a density integrates to 1. Your normalization makes this true for $G(x)$, so it describes the distribution of a random variable on $[0,T]$. I added some additional commentary to that effect to my answer. $\endgroup$
    – Annika
    Feb 1, 2022 at 16:09

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