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Statistics help needed

  1. Nov 17, 2005 #1
    Hey guys I've got several questions about statistics.

    Here's the first one.

    1. Suppose X is a discrete random variable that takes on the three values x1, x2, x3 with probabilities p1, p2, p3 respectively. Describe how you could generate a random sample from X if all you had access to were a list of numbers generated at random from the interval [0, 1].

    2. Suppose X is a continuous random variable with c.d.f. FX(x) = P(X  x). To make things easier, suppose further that X takes on values in an interval [a, b] (do allow for a to be −1 and b to be +1).

    Let Y be the random variable defined by Y = FX(X). This looks strange, but is perfectly
    valid since FX is just a function, and you are allowed to take functions of random variables.

    Your problem: show that Y distributed U[0, 1].

    Method: calculate the c.d.f. Y , FY (y) = P(Y  y), for all real y. Replace Y with FX(X),
    and consider when you can take the inverse of FX.
    Taking the inverse of FX is not always possible, in particular, for x < a and x > b. Consider
    those cases separately.
    You will find that the c.d.f. of Y is 0 for y < 0, y for 0  y  1, and 1 for y > 1, so indeed
    Y distributed U[0, 1].
    An important application of this fact is that if u is a random selection from the interval [0, 1], F−1 X (u) is a random selection from X.
     
  2. jcsd
  3. Nov 17, 2005 #2

    Tom Mattson

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    Regarding the part in blue:

    My browser is showing that as "X square x", as in there is a square-shaped symbol between the X and the x. Is that what you meant? Does anyone else see it differently? Your post is riddled with those squares.
     
  4. Nov 17, 2005 #3
    Concerning part 1. Find the pdf of X and call it f(t). Consider the
    integral int(f(t), t,0,x)=y, where y is one of the numbers generated randomly from (0,1). Solve for x in the above integral. This will generate a quasi-random sequence of samplings of X.
     
  5. Nov 20, 2005 #4
    P(X  x) = P (X <= x)
    and P(Y  y) = P (Y <= y)

    and the other two are strict inequalities <

    Sorry for the confusion.
     
  6. Nov 20, 2005 #5
    hmm..what do you mean by int(f(t), t,0,x)=y...??

    o to x is the limits of integration?
     
  7. Nov 20, 2005 #6
    My familiarity with part 1. comes from Monte Carlo Methods for Integration, in which a definite integral gets approximated by the expectation value of a random variable (who's pdf matches the argument of the integral). In order to sample the random variable, call it x, one creates the equation int(pdf(x),x,o,y)=z, where pdf(x) is the probability density function of the random variable x, y is the variable to be solved for, and z is a random sequence chosen from an arbitrary distribution. As you can see, my example deals with the continuous case, but I would imagine that the discrete case, such as yours, may be dealt with in the same manner.
     
  8. Nov 26, 2005 #7
    Whoa..that's a little hard for me to understand. i don't think i've learn that yet.
     
  9. Nov 26, 2005 #8
    And here's another question:

    2. Suppose X is a discrete random variable that takes on values from {1, 2, 3, . . .} with probabilities
    {p1, p2, p3, . . .}. If u is a number selected at random from [0, 1] explain why

    min {sum from i=1 to n of p subscript i >= u}

    can be considered as a random selection from X.
     
  10. Nov 26, 2005 #9

    EnumaElish

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    Don't you have to know according to which random distribution that list of numbers was generated? I think at the very least you have to assume that you know their distribution even if you don't assume what it is (e.g. uniform).
     
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