Find the cumulative probability function given a probability density function

Question

I am reading a very basic probability introduction. It's mostly a good text, but some small minutiae are often glazed over that someone with a more mathematical background might otherwise understand.

Given a probability mass function $f(w)$ that represents the temperatures you might encounter on a picnic: $$ f(w)= \begin{cases} \frac{w-40}{625}, & \text{if } 40 \leq w \leq 65 ;\\ \frac{90-w}{625}, & \text{if } 65 \leq w \leq 90 ;\\ 0, & \text{otherwise} \end{cases} $$

the cumulative distribution function, $F$, is defined as:

$$ F(w)=\int_{-\infty}^w f(u)dx $$ $$ = \begin{cases} 0, &\text{if } w < 40 ;\\ \frac{w^2-80w+1600}{1250}, & \text{if } 40 \leq w \leq 65 ;\\ \frac{180w-w^2-6850}{1250}, & \text{if } 65 \leq w \leq 90 ;\\ 1, & \text{otherwise} \end{cases} $$

Questions:

1. What is the significance of capital $F$ for the cumulative distribution function as opposed to lower case $f$ for the probability mass function? Is that simply customary for defining a cumulative distribution function?
2. Why do we need a new function, $f(u)$, for the cumulative distribution function? Earlier in the text, it was demonstrated how to find the probability that the temperature would be $\leq55.2$ by:

$$ \int_{40}^{55.2} f(w) dw = \int_{40}^{55.2} \frac{w-40}{625}dw $$ $$ =\frac{w^2-80w}{1250}\Big|_{40}^{55.2} \ $$ $$ =\frac{55.2^2-80\times55.2-40^2+80\times40}{1250} \ $$ $$ =0.185 \text{ (rounded to 3 d.p.)} \ $$ Is this not just a cumulative probability function from $-\infty$ to $55.2$? If so, then why is a new cumulative probability defined as above?

3. How did the author calculate the cumulative probability function? I understand how:

$$ \int_{40}^{55.2} f(w) dw = \frac{w^2-80w}{1250}\Big|_{40}^{55.2} $$ but where does:

$$ \frac{w^2-80w+1600}{1250} $$ come from?

Reference:

Tilman M. Davies. 2015. The Book of R: A First Course in Programming and Statistics (1st ed.). No Starch Press, San Francisco, CA, USA.

Edit:

I can arrive by hand at $$ \int\frac{w-40}{625}dw = \frac{w^2-80w}{1250} + C $$ and $$ \int\frac{90-w}{625}dw = \frac{180w-w^2}{1250} + K $$

Where does the $1600$ and the $6850$ in the numerators of the two functions below come from?

$$ \frac{w^2-80w + 1600}{1250}\text{if }40\leq w \leq 65; $$ and $$ \frac{180w-w^2-6850}{1250}\text{if }65 < w \leq 90; $$

I'm sorry that I am also asking for a quick explanation on some basic integral math.

Answer 1

Partial: I will only answer how they got $F(w)$ for $40 \leq w \leq 65$. $F(w)=\int_{40}^{w} f(u) du =\int_{40}^{w} \frac {u-40} {625}du =(\{u^{2}/2 -40u\}_{u=w} - \{u^{2} /2-40u\}_{u=40})/625$ =$\frac {w^{2}-80w-2(40^{2}/2-1600)} {1250}$=$\frac {w^{2}-80w+1600} {1250}$

Answer 2

1) You must be familiar with statistical concepts of "frequency" (denoted by $f$) and "cumulative frequncy" (denoted by $F$). The concepts of "probability density function" and "cumulative probability density function" are extensions of those statistical concepts to probability (percentage, proportion) concepts.

2) You are right, PDF is integrated to get CDF. The author did the same: $$\int \frac{w-40}{625}dw=\frac{(w-40)^2}{1250}+C=\frac{w^2-80w+1600}{1250}+C.$$

Answer 3

1. This a naming convention. It is useful, however to denote the cdf and the pdf of the same random variable with the same letter -- yet the two notations have to be different as the functions are different.

2.

The cdf. exists always, the pdf does not exist necessarily. However it is easier to use the pdf.

3. $$ \frac{w^2-80w+1600}{1250} $$

comes from $$\int_{40}^wf (x)dx=P (40 <X\leq w) =F (w)-F (40)$$

where $40 <w\leq 65$ and $X $ is the random variable about whose cdf and pdf we are talking.