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I'm studying stochastic calculus now. And I found in my textbook that strictly review elementary probability theory.

But I suddenly confused the notion of uniform distribution.

I thought when we see uniform distribution on the 2 dimensional space, it just flat line on X-Y graph.

like below picture(from wikipedia), uniform distribution has the function $P[a,b] = \frac{1}{b-a}$

enter image description here

However In my textbook, it just change it's function from $P$ to $P_{tilda}$, I mean from $b-a$ to $b^2 - a^2$. enter image description here

I think $P[a,b] = \frac{1}{b-a} ≠ b-a ≠ b^2 - a^2$. So both equation about $P[a,b]$ in the textbook are not uniform distribution. But the textbook said, $b-a$ is uniform, $b^2 - a^2$ is no longer has the uniform distribution.

So I'm now confused of what is the real meaning of 'uniform' in probability..!

In short, What is the meaning and reason of the sentence in the above picture : "Under P tilda, the random variable X no longer has the uniform distribution"?

*the text book is stochastic calculus for finance Ⅱ by shreve

user13232877
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    TL;DR. Apparently you are confusing the density $f$ of the uniform distribution on $[a,b]$ with its integral. The density must be $f(x)=1/(b-a)$ for the integral over $[a,b]$ to be one (probability distribution). The example 1.2.4 from that book considers a uniform distribution on $\color{red}{[0,1]}$. Now it seems you are confusing $0$ with $a$ and $1$ with $b$. Note every $a$ in different references means the same thing! – Kurt G. Apr 29 '22 at 06:02
  • You mean that 1. probability is integral of density function $f$ 2. So in the textbook, if $P[a,b] = b-a$ then it means it's density function is constant "1". Therefore it is uniform distribution. 3. However, if $P[a,b] = b^2-a^2$ then it means it's density function is $x$; considering that $(∫^b_af(x)dx = Probability)$. Therefore it is not uniform distribution because density function is monotonic increasing. Am I right? – user13232877 Apr 29 '22 at 06:28
  • .You are right. – Kurt G. Apr 29 '22 at 06:44

1 Answers1

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I should not have confused about the definition.

Definition : $P(X∈[a,b]) = ∫^b_af(t)dt$

As Kurt.G mentioned, Probability is integral of density function : $f(x)$

Flat line of uniform distribution on the graph means that we deal with "density funciton".

For example, Let's say we are focusing the interval $0 ≤ a ≤ b ≤ 1$. To be one(probability's property) when integrated, it needs to be $f(x) = \frac{1}{b-a}$

It means $∫^b_af(t)dt = ∫^b_a\frac{1}{b-a}dt = \frac{1}{b-a}[t]^b_a = \frac{b-a}{b-a} = 1$

So the textbook tells us about "Probability" not "density" function.

Again, $P[a,b] = ∫^b_af(t)dt$.

To be $P[a,b] = b-a$, $f(t)$ needs to be constants. Just "1". Then it can translated that now we have flat density function, meaning uniform distribution.

Also To be $P[a,b] = b^2 - a^2 $, $f(t)$ needs to have first order term with "$t$" It means density function is increasing, therefore it is not uniform distribution.

user13232877
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