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Functions of Continuous random variables

  1. Nov 11, 2005 #1
    I have been working on this problem and can't seem to get the answer.

    X is a continuous random variable with a proabaility density function:

    f(x) = 1/4 if -2<=x<=2
    0 other wise

    Let Y=1/X. Then P(Y<=1/2) = ?

    This is how I approached the problem:


    Taking the intergral of f(x) with limits of integration -2,2.

    My answer is 1.

    However the answer given is 1/2.

    Any ideas.
  2. jcsd
  3. Mar 30, 2006 #2

    Hmm.. Since it is given in the question that x runs from -2 to 2 only, how can [tex] P(X\geq2) [/tex] be 1? Shouldn't it be 0?

    Well, your working [tex] P(Y\leq \frac{1}{2}) = P(\frac{1}{X}\leq \frac{1}{2}) [/tex] is correct. The error lies in your subsequent statement "which is in turn equal to [tex] P(X\geq2) [/tex]"

    You see, since x runs from -2 to 2, it can take both positive and negative values. So, should we treat x as positive or negative, knowing that the subsequent working will be affected by our decision? (i.e. If x is negative, then we will need to change the inequality sign when cross-multiplying)

    To solve this problem, we multiply the inequality by [tex] x^2 [/tex], since we know for sure that this expression is positive (x cannot be equal to zero if y is to be real)

    The resulting inequality will be quadratic in nature, and some algebraic manipulation should get you the desired probability of 0.5.

    All the best!
    Last edited: Mar 30, 2006
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