# Distribution function and Densities

• Oct 11th 2009, 04:52 AM
ryanhorne
Distribution function and Densities
These question seemed simple to me, however I believe I am getting confused between distribution functions and densities.

Question

Find the distribution function, and hence the density, of (1/x)

Any help would be great. The reason why im confused is because i thought 1/x was already a density, so why would it need to be found?

Thanks
• Oct 11th 2009, 06:34 PM
matheagle
Do you have an orginal distribution?
I can tranform any rv via 1/x, if I know what distributon we are starting with.
In other words.
Give me the density of X and I will give you the density of Y=1/X.
• Oct 12th 2009, 07:45 AM
ryanhorne
Yes it is a random variable, and I have been told that the range is from

0 < x < 2

Does this help?
• Oct 12th 2009, 07:48 AM
matheagle
nope
how about an f(x) or an F(x)?
• Oct 12th 2009, 10:33 AM
ryanhorne
OK, here is the complete question:-

X has density x/2 over 0 < x < 2, and 0 elsewhere. Sketch its density and distribution function. Find the distribution function, and hence the density, of X/2, e^X, 1/X, and |X - 1|.

So all I know is that it has something to do with random variables. Please dont explain all of them as that is time consuming, if you can do one then Im sure I can work out the rest with the same technique.
• Oct 12th 2009, 12:08 PM
matheagle
Quote:

Originally Posted by ryanhorne
OK, here is the complete question:-

X has density x/2 over 0 < x < 2, and 0 elsewhere. Sketch its density and distribution function. Find the distribution function, and hence the density, of X/2, e^X, 1/X, and |X - 1|.

So all I know is that it has something to do with random variables. Please dont explain all of them as that is time consuming, if you can do one then Im sure I can work out the rest with the same technique.

that is just calculus one

Let Y=1/X

the density of Y is

$\displaystyle f_Y(y)=f_X(x)/y^2$ since we need the absolute value around the $\displaystyle -y^{-2}$

Thus $\displaystyle f_Y(y)={1\over 2y}y^{-2}={1\over 2y^3}$

for y>1/2
• Oct 13th 2009, 08:44 AM
ryanhorne
Just had a thought, do you reckon the final answer should have been negative?
• Oct 13th 2009, 09:08 AM
matheagle
Quote:

Originally Posted by ryanhorne
Just had a thought, do you reckon the final answer should have been negative?

probabilities are never negative
same with probability functions
• Oct 14th 2009, 05:53 AM
ryanhorne
haha, good point. (Clapping)