# Thread: Power Distribution

1. ## Power Distribution

I have an example from a book that I dont quite understand, where they talk about drawing random samples from some domain, using a Power Distribution. With the domain being from 0-1.

Now, what I understand this to mean is that with a uniform probabilty distribution, the chances of picking any number from the domain are equal, so the pdf is just 1/domain. But with a Power Distribution pdf, then the probabilty of picking a number from the domain gets higher as the variable increases in value.

ie if i have the exponent (n), for this power distribution set to say 3,

the probability of picking 0.33 is 0.33^3
the probability of picking 0.78 is 0.78^3

But the pdf, needs to sum to 1 over the domain, so you need to add a normalization constant to do this, (c)

pdf = cx^n

With the restriction that the pdf has to sum over one, you integrate the above from 0-1 and work out the constant to be

c = n + 1

so now the pdf is

pdf(x) = (n + 1)x^n

But if i try this with examples, i get probabilities over 1.

n = 8

pdf(0.6) = (8+1)0.8^8 = 1.50
pdf(0.8) = (8+1)0.6^8 = 3.54
pdf(0.99) = (8+1)0.99^8 = 8.30

2. With no replies, I assume I havent explained my problem very well. Below is the exact example in question.

Ignoring the stuff about the CDF for the time being, Im wanting to test the pdf that they calculated.

with n = to 8, why do I have probabilties over 1? the values im using are in the correct domain of 0-1

pdf(0.6) = (8+1)0.6^8 = 1.50
pdf(0.8) = (8+1)0.8^8 = 3.54
pdf(0.99) = (8+1)0.99^8 = 8.30

3. I think your confusion is with the interpretation of a pdf.

the pdf f(x) does not tell you the probability that x takes on a certain value. For the purposes of high school statistics, the probability of x taking any particular value exactly can be thought of as zero for all continuous distributions.

The way to work out probabilities for a continuous distribution is to integrate the PDF to get the CDF, you can use that to work out probabilities like P(X < 0.8).

4. Originally Posted by SpringFan25
I think your confusion is with the interpretation of a pdf.

the pdf f(x) does not tell you the probability that x takes on a certain value. For the purposes of high school statistics, the probability of x taking any particular value exactly can be thought of as zero for all continuous distributions.

The way to work out probabilities for a continuous distribution is to integrate the PDF to get the CDF, you can use that to work out probabilities like P(X < 0.8).

So what exactly does pdf(x) = (n + 1)x^n say? (with a continuous function)

I thought that if you're using a function that generates continuous values over some range, say 0 - 50, with distribution x^n, then the probability of a getting a larger number in that range is higher than getting a lower number due to the ^n, and that you can work out the probability of getting a particular number in the specified range using the above pdf.

Whilst it makes sense that the probabilty for some number over a continuous function would be 0, im wondering what the pdf for a continuous function actually is telling you?

So if I do want to compute the probabilty for some value i have to instead generate a very small interval around it?

ie

CDF(x+SMALLVAUE) - CDF(x-SMALLVAUE)

?

5. So what exactly does pdf(x) = (n + 1)x^n say? (with a continuous function)
The PDF tells you the rate of change of the CDF. It can be shown that the pdf is proportionate to the probability of x taking a particular value. example, if:
f(2) = 1
f(3) = 2

then x=3 is twice as likely to occur as x=2. However it doesn't tell you the level of the probabilities, just their relative size.

So if I do want to compute the probabilty for some value i have to instead generate a very small interval around it?

ie

CDF(x+SMALLVAUE) - CDF(x-SMALLVAUE)
This will tell you that chance of x falling in the interval