# Probability density function of a random variable.

1. Apr 26, 2010

### Fuquan22

1. The problem statement, all variables and given/known data
Let X be a posative random variable with probability density function f(x). Define the random variable Y by Y = X^2. What is the probability density function of Y? Also, find the density function of the random variable W = V^2 if V is a number chosen at random from the interval (-a,a) with a>0.

2. Relevant equations

3. The attempt at a solution
I know the answer to the first part is f(sqrt(y))/2sqrt(y) and the answer to the second part is 1/(2a*sqrt(w)) but i have no idea how to get to these answers. I have looked through my book and notes many times and i'm still lost. Any help would be very much appreciated

2. Apr 26, 2010

### zachzach

What is the probability density function? To solve think change of variables and integration.

3. Apr 26, 2010

### Fuquan22

What do I integrate? The only way i could see how to do it was to derive but i still didn't know what to derive and how to plug it in.

4. Apr 26, 2010

### zachzach

Can you define what a probability density function is in general terms?

5. Apr 26, 2010

### Fuquan22

A probability density function of a continuous random variable is a function that describes the relative likelihood for this random variable to occur at a given point in the observation space.

6. Apr 26, 2010

### zachzach

7. Apr 26, 2010

### zachzach

You do not use integration. I did Google search and the formula is in many places. Basically the probability contained in a differential area must be equal in each of the cases.

$$|f(x)dx| = |f(y)dy|$$

8. Apr 26, 2010

### Fuquan22

ok but how does that get me to my answer?

9. Apr 26, 2010

### zachzach

10. Apr 27, 2010

### Fuquan22

Ok thanks, how does it apply to the second part of the question then? I figured out the first part though.

11. Apr 27, 2010

### Legendre

My course teaches us to derive from basic definitions. Not sure if its the best way but its the only way I know. :p

(Cumulative) Distribution Function of Y
= F(y)
= P(Y < or = y)
= P(X^2 < or = y)
= P( -sqrt y < or = X < or = sqrt y)
= [ 1 - P(x < -sqrt y) ] + P(x < or = sqrt y)
= 1 - G(-sqrt Y) + G(sqrt Y)

Where G is the Distribution Function of X.

To get to the Density of Y, we need to differentiate F(y). Remember that G contains a function of Y so you need the chain rule.

For W = V^2, V has a uniform distribution on (-a,a). Everything is the same except "Y" becomes "W" and "X" becomes "V". Use the density of V accordingly.

P.S.
your answer to the first part appears to be wrong. which implies the 2nd part is wrong too.

Last edited: Apr 27, 2010
12. Apr 27, 2010

### Fuquan22

Answers to the first and second part aren't wrong, those are the answers in the back of my book.

13. Apr 27, 2010

### Legendre

Have you looked at my suggested method? After you differentiate the cumulative distribution function F(y) to get the probability density f(y), the answer is different from yours. Maybe there is a printing error. (happens all the time with textbooks)

14. Apr 27, 2010

### Fuquan22

I did look at your method Legendre, and i think i just do the integral of f'(sqrt(y)), which by the chain rule gives me f(y)/(2sqrt(y)) which is the answer i need.

15. Apr 27, 2010

### Legendre

You are right. The answer is correct. Sorry about the confusion!

Cumulative Dist. = F(y) = 1 - G(-sqrt Y) + G(sqrt Y)

Density = f(y) = d/dy [F(y)] = (1/2sqrt(y))[ g(-sqrt Y) + g(sqrt Y)]
where g is the density of X.

g(-sqrt Y) = 0 since X is a positive random variable.

So, f(y) = (1/2sqrt(y))(g(sqrt Y)).

P.S.
I apologize for mixing up the notation! I know f is supposed to be the density of X in your question but for the sake of being consistent with my previous post, I let f be the density of Y and g be the density of X instead. Sorry!

16. Apr 27, 2010

### Fuquan22

Cool thanks. Do you know how I would get the second part though. I know its got to be similar but in what step do i plug in the a. And why is it 1 over the answer rather than f(sqrt(W))? Thanks for the help

17. Apr 27, 2010

### Legendre

You're welcomed man! Getting stuck at maths sucks! I got stuck for days recently and this forum helped me out, so I know how valuable maths help can be!

W = V^2

1) V has a continuous uniform density function over (-a,a) for a>0. Let g(v) be the density function for V. g(v) = 1/2a. The formula for density of continuous uniform distributions on interval (a,b) is "1/(b-a)".

2) V is not a positive random variable because V is a number chosen from (-a,a). V < 0 for some outcomes.

3) Considering the previous point, we need to re-derive the equation f(y) from our previous part. I am sticking to my notation for consistency. Apologies if its confusing.

F(y) = 1 - G(-sqrtY) + G(sqrtY)

Then f(y) = (1/2sqrt(y))[ g(-sqrt Y) + g(sqrt Y)

4) We know g(v) = 1/2a for all v, so we plug it in and replace all y with w.

f(y) = (1/2sqrt(w))(1/2a + 1/2a) = (1/2sqrt(w))(1/a) = 1/(2a*sqrt(w)) as required.

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