# I need help making up a skewed PDF

• Mar 7th 2011, 03:40 PM
xamius
I need help making up a skewed PDF
Basically I need to make up a PDF that is skewed enough that it wont really fit the CLT unless I have a large number of observations. I have no idea how to just make one up (I cant use anything with a name like Chi squared, poisson, etc.)

I know obviously the properties of a PDF (it must have all positive values, the integral has to equal 1, etc.)

I just need some help on how to start.

• Mar 7th 2011, 09:56 PM
Sambit
You may construct a discrete probability distribution with more frequencies on the left than on the right.
• Mar 8th 2011, 08:18 AM
xamius
the problem is construct out of what? I have never had to do anything but use a given PDF to find this probability etc. I dont know how to create a PDF from scratch.
• Mar 8th 2011, 08:38 AM
theodds
Consider densities of the form:

$\displaystyle f(x) = \theta x ^{-\theta - 1}I(x \ge 1)$

(Note: I came up with this on my own, but after a few minutes I realized this is a Pareto distribution. So, it has a name, but it is very easy to come up with on your own so I would think it still counts since the Pareto isn't usually covered extensively).

As far as coming up with your own pdfs: every function that is positive and has a finite integral - i.e. $\displaystyle \int_{-\infty}^ \infty f(x) \ dx < \infty$ - can be turned into a pdf by dividing by the value of the integral. So, pick a function that integrates and then normalize it to get a pdf.
• Mar 8th 2011, 08:42 AM
xamius
okay first off what is I? Second, when you say I came up with this on my own, the question is how? What made you think of that?
• Mar 8th 2011, 08:50 AM
theodds
$\displaystyle I(\cdot)$ is the indicator function. It is equal to 1 if the statement inside is true, and 0 otherwise. It is a more compact way of writing

$\displaystyle f(x) = \begin{cases} \theta x ^ {-\theta - 1} \qquad & x \ge 1 \\ 0 \qquad & x < 1 \end{cases}$

I came up with it by thinking of a function that was integrable on the whole real line, but did not satisfy $\displaystyle \mbox{E} X ^ p < \infty$ for all p. So, for appropriate choice of theta, the mean, variance, skewness, kurtosis, etc. may or may not exist. This gives me a good deal of control over how long it takes the central limit theorem to work. Of course, it probably wouldn't be reasonable to expect a student to think of all those things at once, but it IS skewed for appropriately chosen theta which is something you could notice (I won't tell you which ones, though, because I should at least leave you some work to do).

A word of caution if you decide to use this: the choice of theta is important. If you choose theta too small the CLT won't apply at all - obviously theta needs to be positive, but that isn't the end of it if you want to claim the CLT holds.
• Mar 8th 2011, 04:11 PM
xamius
Um so in terms of coming up with my own, when you say divide it by the value of its integral, so you take the function Y, then integrate it from -infinity to inifinity and that is the pdf?
• Mar 8th 2011, 04:16 PM
xamius
for example how would you make e^x a pdf?

or could you not since the integral is e^x?
• Mar 8th 2011, 04:46 PM
theodds
Quote:

Originally Posted by xamius
for example how would you make e^x a pdf?

or could you not since the integral is e^x?

To turn $\displaystyle f(x)$ into a pdf, just take

$\displaystyle \displaystyle p(x) = \frac{f(x)}{\int_{-\infty} ^ \infty f(x) \ dx}.$

This doesn't work for e^x because $\displaystyle \int_{-\infty}^\infty e^x \ dx = \infty$.
• Mar 8th 2011, 06:10 PM
xamius
so for example the pdf of the function 2x^2 is 3/x?
• Mar 8th 2011, 06:29 PM
theodds
It's a definite integral...