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

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

http://i51.tinypic.com/10x4rpl.jpg

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

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).

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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).

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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)

?

Quote:

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

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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.

Quote:

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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)

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This will tell you that chance of x falling in the interval