# Math Help - Probability density function

1. ## Probability density function

Continuous random variable X has probability density function defined as

f(x)= 1/4 , -1<x<3

=0 , otherwise

Continuous random variable Y is defined by Y=X^2

(1) Find P(X>2) given that X>0

Ans: (1)(1/4)=1/4

(2) Find G(y), the cummulative distribution function of Y

f(y)= 1/4 , 1<X<9

= 0, otherwise

Then integrate to get the cdf.

G(y)= 0 , Y<=1

= 1/4 y , 1<Y<9

=1 , Y>=9

(3) Hence find the probability density function of Y.

Then now differentiate the cdf in (2) to get the pdf

Part(2) doesn't seem right so do part(3).

2. Originally Posted by hooke
Continuous random variable X has probability density function defined as

f(x)= 1/4 , -1<x<3

=0 , otherwise

Continuous random variable Y is defined by Y=X^2

(1) Find P(X>2) given that X>0

Ans: (1)(1/4)=1/4

(2) Find G(y), the cummulative distribution function of Y

f(y)= 1/4 , 1<X<9

= 0, otherwise

Then integrate to get the cdf.

G(y)= 0 , Y<=1

= 1/4 y , 1<Y<9

=1 , Y>=9

(3) Hence find the probability density function of Y.

Then now differentiate the cdf in (2) to get the pdf

Part(2) doesn't seem right so do part(3).
(2) cdf $= G(y) = \Pr(Y \leq y) = \Pr(X^2 \leq y) = \Pr(-y \leq X \leq y)$. Calculate this probability using the given pdf for X.

3. Originally Posted by mr fantastic
(2) cdf $= G(y) = \Pr(Y \leq y) = \Pr(X^2 \leq y) = \Pr(-y \leq X \leq y)$. Calculate this probability using the given pdf for X.
Thanks Mr Fantastic,

For this step, $\Pr(X^2 \leq y) = \Pr(-y \leq X \leq y)$,

is X^2=|X|? But $3^2\neq |3|$ so how did this step happen?

And i do not know y so how do i find the probability?

4. Originally Posted by hooke
Thanks Mr Fantastic,

For this step, $\Pr(X^2 \leq y) = \Pr(-y \leq X \leq y)$,

is X^2=|X|? But $3^2\neq |3|$ so how did this step happen?

And i do not know y so how do i find the probability?
$X^2 \leq y \Rightarrow -y \leq X \leq y$ is a basic inequality.

You're meant to integrate the given pdf for X between -y and y to get an expression with y. This expression is the cdf. You should review your class notes or textbook for similar examples.

5. Originally Posted by mr fantastic
$X^2 \leq y \Rightarrow -y \leq X \leq y$ is a basic inequality.

You're meant to integrate the given pdf for X between -y and y to get an expression with y. This expression is the cdf. You should review your class notes or textbook for similar examples.
ok

the probability of P(-y<=X<=y)=1/2 y

so the cdf is

G(y)= 0 , x,-y

= 1/2 y , -y<=X<=y

=1, x>y

Am i correct?

6. Originally Posted by hooke
ok

the probability of P(-y<=X<=y)=1/2 y

so the cdf is

G(y)= 0 , x,-y

= 1/2 y , -y<=X<=y

=1, x>y

Am i correct?
No. You're probably trying to punch above your weight at the moment. Nevertheless, consider carefully the following:

If $-1 \leq x \leq 3$ and $Y = X^2$ then clearly $0 \leq y \leq 9$ (draw a graph to see this). There are therefore three cases you have to consider:

Case 1: $y < 0$. $G(y) = 0$.

Case 2: $0 \leq y < 1$. $\displaystyle G(y) = \int_{-\sqrt{y}}^{\sqrt{y}} f(x) \, dx$ (why?)

Case 3: $y \geq 1$. $\displaystyle G(y) = \int_{1}^{\sqrt{y}} f(x) \, dx$ (why?)

7. Originally Posted by mr fantastic
No. You're probably trying to punch above your weight at the moment. Nevertheless, consider carefully the following:

If $-1 \leq x \leq 3$ and $Y = X^2$ then clearly $0 \leq y \leq 9$ (draw a graph to see this). There are therefore three cases you have to consider:

Case 1: $y < 0$. $G(y) = 0$.

Case 2: $0 \leq y < 1$. $\displaystyle G(y) = \int_{-\sqrt{y}}^{\sqrt{y}} f(x) \, dx$ (why?)

Case 3: $y \geq 1$. $\displaystyle G(y) = \int_{-1}^{\sqrt{y}} f(x) \, dx$ (why?)
Sorry this is not the kind of cdf problems i usually encounter so pls be patient with me.

How did you get the domain of y for case (2)? Maybe that will help me figure out why integrate from -root(y) to root(y).

8. Originally Posted by hooke
Sorry this is not the kind of cdf problems i usually encounter so pls be patient with me.

How did you get the domain of y for case (2)? Maybe that will help me figure out why integrate from -root(y) to root(y).
Think about this: f(x) = 0 for X < -1. So what would happen in Case 2 if you did have y > 1 ....

9. Originally Posted by mr fantastic
Think about this: f(x) = 0 for X < -1. So what would happen in Case 2 if you did have y > 1 ....
Is it right to say that

f(x^2)=f(y)=1/4 for 0<=Y<=9 ?

Sorry but i don't get your point. Wouldn't that be case 3 ?

10. Originally Posted by hooke
Is it right to say that

f(x^2)=f(y)=1/4 for 0<=Y<=9 ?

Sorry but i don't get your point. Wouldn't that be case 3 ?
No.

My point is that if y > 1 then $-\sqrt{y} < -1$ and so f(x) and hence the integral of f(x) is equal to 0, which means that the integration will only be non-zero for 0 < y < 1.

When y > 1, the integral is only non-zero for x > -1, that is, Case 3 is required (and I just realised that I made a typo in Case 3, the lower integral terminal shuold be -1 NOT 1. I have edited that post).

At this point, I think the best thing for you to do is print out this thread and talk it through one-on-one with your instructor.