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Probability of continuous random variables

  1. Nov 16, 2007 #1
    1. The problem statement, all variables and given/known data
    A random variable has distribution function F(z) = P(y<= z) given by (this is a piecewise function)

    f(z) =
    0 if z < -1
    1/2 if -1 <= z < 1
    1/2 + 1/4(z-1 if 1 <= z < 2
    1 if 2 <= z

    What is P(Y = 2)?

    Find all the numbers t with the property that both P(Y <= t) >= 1/2 and P(Y >= t) >= 1/2

    2. Relevant equations

    3. The attempt at a solution

    For the P(Y=2) I integrated at the point 2 plus and minus epsilon and came up with 1/4z - 0 where z =2. Thus, 1/2. My concern is that this problem has a lot of constants Thus, I would expect P(Y=x) to equal 0 everywhere but in [1,2). Then I have no idea how to find the median of the distribution function. Sorry, if these are easy questions. The class is being taught without a book and I'm afraid I'm not used to that.
  2. jcsd
  3. Nov 16, 2007 #2


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    I don't understand what you're doing -- could you spell it out? The answer is certainly wrong.

    Not just a lot of flat-lines in the graph, but it's mostly continuous! P(Y=a) can only be positive if the graph of the cumulative distribution function has a discontinuity. (right?)

    Do you know how to find P(Y<=t)?
  4. Nov 16, 2007 #3
    well, no expert here, but to find P(Y=2) shouldn't you integrate f(z) over (-infinity,2]?
  5. Nov 16, 2007 #4
    F(z) = P(y<= z)

    what exactly is "y" here?

    the CDF is supposed to be F(z) = P[Z <= z]
  6. Nov 17, 2007 #5


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    He said that f was the cumulative distribution, not the probability density.
  7. Nov 17, 2007 #6


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    This is impossible. The integral of a probability density function over its entire domain must be 1. You obviously can't have "f(z)= 1 if 2<= z". It also cannot be a cumulative probabililty distribution which was my next guess. I have no idea what is intended here.

    No, that would give P(Y[itex]\le 2[/itex]). With a continuous probability density, the probobability that Y is any specific number is 0.
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