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Random variable transform.

  1. Jun 26, 2011 #1
    1. The problem statement, all variables and given/known data

    part iv confuse me,especially the limits for y
    please look to my answer for this part and comment

    2. Relevant equations



    3. The attempt at a solution

    i) I got c = 1/3
    ii) P(X^2 >=1)=P(X>=1) + P(X<= -1) = 7/9
    iii) P(X-1>=-1/4) = P(X>= -1/4+1)=37/576

    iv) we find cdf of X
    F(x) = integral from ( -2 to x ) of f(x) = int (1/3x^2) = (1/9)[ x^3 + 8]

    the cdf of Y:
    P(Y<=y) = P(-X<=y) = P(X>=-y) = 1- P(X<=-y) = 1-F(-y) = 1-(1/9)[- y^3 + 8]
    the limits of Y ??????????

    pdf of Y: f(y) = dF(y)/dy = (1/3) y^2

    ------------------------------------------------
    the correct answer:
    pdf of Y : f(y) = 1/9 * impulse(y) , y =0
    = 1/3 y^2 , 0<y<2
    why?????????????????????????????????
     

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  3. Jun 26, 2011 #2
    Welcome to PF, windowsxp!

    You need to be more careful about the conditions on x when deriving the cdf. You should be getting piecewise functions.

    For instance, your result for the cdf of X is incorrect. If I take x = 1000, your formula gives [itex] F_X(x) = \frac{1000^3 + 8}{9} = 111,111,112 [/itex], which can't possibly be right. Similarly, if I choose a large negative number, your formula yields a negative probability, which is impossible. However, for values between -2 and 1, your formula is right. Just make sure you look at the piecewise definition of the density function when calculating the cdf. Your result have the form
    [tex]
    F_X(x)=
    \begin{cases}
    ? & \text{if} \quad x>1,
    \\
    \frac{x^3 + 8}{9} & \text{if} \quad -2 \leq x \leq 1,
    \\
    ? & \text{if} \quad x < -2
    \end{cases}
    [/tex]

    and you should be able to fill in the question marks without too much difficulty.

    Similarly for the rest of part iv), you need to take into consideration the fact that Y is also defined as a piecewise function. So there are two cases, [itex] X \leq 0[/itex] and [itex] X > 0 [/itex]. The strange thing is that [itex] \mathbf{P}\{Y = 0\}[/itex] is nonzero, which is why the correct answer for the pdf has that extra condition.
     
  4. Jun 26, 2011 #3
    thanx a lot for your comments.

    the cdf of X is
    F_X(x)=
    \begin{cases}
    0 & \text{if} \quad x>1,
    \\
    \frac{x^3 + 8}{9} & \text{if} \quad -2 \leq x \leq 1,
    \\
    1 & \text{if} \quad x < -2
    \end{cases}
    [/tex]
    right!!!

    Y is defined by two intervals
    when X < 0 , Y= -X
    from the cdf of X, F(x) = (1/9)[ x^3 + 8] , -2<x< 0
    since -2<x< 0 , 0< y< 2
    P(Y<=y) = P(-X<=y) = P(X>=-y) = 1- P(X<=-y) = 1-F(-y) = 1-(1/9)[- y^3 + 8]
    pdf of Y: f(y) = dF(y)/dy = (1/3) y^2 ,0< y< 2

    when X >0 , Y=0
    from the cdf of X, F(x) = (1/9)[ x^3 + 8] , 0<x< 1
    .....................?
     
  5. Jun 26, 2011 #4
    Almost, you just mixed up the 0 and the 1. But you have the right idea! :smile:
    The problem is, if you assume [itex] X \leq 0 [/itex], then you're actually calculating a conditional probability. It's clearer to write the event [itex]\{Y \leq x\} [/itex] as the disjoint union [itex] \{Y \leq x \; \text{and} \; X \leq 0\} \cup \{Y \leq x \; \text{and} \; X > 0\} [/itex]. Then
    [tex]
    \mathbf{P}\{Y \leq x\} = \mathbf{P}\{Y \leq x \; \text{and} \; X \leq 0\} + \mathbf{P}\{Y \leq x \; \text{and} \; X > 0\}.
    [/tex]
    Now let's look at the first term:
    [tex]
    \mathbf{P}\{Y \leq x \; \text{and} \; X \leq 0\} = \mathbf{P}\{-X \leq x \; \text{and} \; X \leq 0\}= \cdots
    [/tex]

    Can you figure out the first term from there? The second term should be easy from the piecewise definition of Y.
     
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