Probability density function

• Sep 3rd 2010, 01:57 AM
hooke
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).
• Sep 3rd 2010, 04:57 AM
mr fantastic
Quote:

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.
• Sep 4th 2010, 01:29 AM
hooke
Quote:

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?
• Sep 4th 2010, 02:10 PM
mr fantastic
Quote:

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.
• Sep 4th 2010, 09:31 PM
hooke
Quote:

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?
• Sep 4th 2010, 09:55 PM
mr fantastic
Quote:

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?)
• Sep 5th 2010, 01:54 AM
hooke
Quote:

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).
• Sep 5th 2010, 04:33 AM
mr fantastic
Quote:

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 ....
• Sep 5th 2010, 07:54 AM
hooke
Quote:

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 ?
• Sep 5th 2010, 12:24 PM
mr fantastic
Quote:

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.