A Seemingly Discrete Distribution Which Is Actually Continuous.

91 Views Asked by At

I have been working on my Statistics recently, and while working through the book I have found a problem I didn't quite understand. The setting is like this:

A box contains good/defective items mixed in thoroughly. If an item drawn is good, the number $1$ is assigned to the drawing; otherwise the number $0$ is assigned. Let $p$ be the probability of drawing a good item at random.

-> So the PDF is:

$ f(x)=P(X=x)= \begin{cases} 1-p&\text{if}\, x= 0\\ p&\text{if}\, x=1\\ \end{cases} $

->I was stuck with the CDF, the manual showed that it is this:

$ F(x)= \begin{cases} 0&\text{if}\, x< 0\\ 1-p&\text{if}\, 0\leq x\lt 1\\ 1&\text{if},x\geq1 \end{cases} $

->My questions are the following:

  1. How is this a continuous distribution? The drawings are discrete right? how come the CDF is for a continuous distribution?

  2. I know the following properties in order that a CDF be classified as continuous:

$\qquad$ a) $F'(x)=f(x)$,

$\qquad$ b) $F(-\infty)=0$,

$\qquad$ c) $F(\infty)=1$

$\qquad$ d) $F(x)$ is defined everywhere in a continuous range.

In regards to the properties a-d, for b,c and d I readily understood what happened where so those are checked for this problem but then what about property a? I just can't seem to find a way to use it to show that the CDF is indeed continuous.

  1. We know $\int_{-\infty}^{x}f(t)dt$ yields the CDF from PMF, in this example how do i use this relation?
1

There are 1 best solutions below

1
On BEST ANSWER

Your random variable is a Bernulli. It has only two probability masses concentred in $X \in \{0;1\}$

The CDF (cumulative distribution function) is the one your textbook showed. As you can see doing a drawing it is not continuous but it is a "steps function".

enter image description here

To get the probability values you have not to derivate $F$ but simply do the difference:

$$p(X=x_0)=F_X(x_0)-F_X(x_0^-)$$

thus, to derive

$$p(X=0)=(1-p)-0=1-p$$

and

$$p(X=1)=1-(1-p)=p$$