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Math Help - Probability density function

  1. #1
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    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).
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    Quote Originally Posted by hooke View Post
    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.
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    Quote Originally Posted by mr fantastic View Post
    (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?
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    Quote Originally Posted by hooke View Post
    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.
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    Quote Originally Posted by mr fantastic View Post
    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?
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    Quote Originally Posted by hooke View Post
    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?)
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    Quote Originally Posted by mr fantastic View Post
    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).
    Last edited by mr fantastic; September 5th 2010 at 12:25 PM.
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    Quote Originally Posted by hooke View Post
    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 ....
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    Quote Originally Posted by mr fantastic View Post
    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 ?
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    Quote Originally Posted by hooke View Post
    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.
    Last edited by mr fantastic; September 5th 2010 at 12:37 PM.
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