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A probability density

  1. Oct 11, 2015 #1
    1. The problem statement, all variables and given/known data
    If the distribution function of a random variable is given by
    F(x) = 1- 1/x^2 for x>1
    F(x) = 0 for x <= 1

    find the probabilities that this random variable will take on a value
    a) less than 3
    b) between 4 and 5

    3. The attempt at a solution
    since they use the capital F i think that means it is a cumulative probability distribution. So in order to find part a) we need to find the cumulative probability that it takes on a value from 3 to infinity and subtract this from 1
    the antiderivative of the function for x>1 is

    x + 1/x

    we need to evaluate this from x=3 to x = infinity
    we have
    ∞ + 1/∞ - (3 + 1/3)
    = ∞ + 0 -3 - 1/3
    = ∞ - 3.333

    I know this can't be right because of the ∞ but I am not sure what I am doing wrong =[ I would appreciate it if someone can't point me in the right direction
  2. jcsd
  3. Oct 11, 2015 #2
    wait a minute. I found in my book that dF(x)/dx = f(x)
    so since F(x) = 1 - 1/x^2

    f(x) = dF(x)/dx = d/dx(1-1/x^2) = 0 - d/dx(x^-2) = -(-2x^-3) = 2x^-3

    so to find the cumulative probability it takes on a value from 3 to ∞ we just plug in
    (1- 1/∞^2) - (1-1/3^2)
    = (1-0) - (1- 1/9)
    = 1 - .888888...
    = .111111...
    but since we want the probability of the value taking on a value less than 3, we subtract this from 1
    so the answer to part a) is .888888
    is this right?
  4. Oct 11, 2015 #3

    Ray Vickson

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    Homework Helper

    Why are you doing it the hard way? You are given the distribution function (= cumulative distribution)
    [tex] F(x) = \begin{cases}
    0 &\text{if} \;x \leq 1 \\
    \displaystyle 1 - \frac{1}{x^2}& \text{if} \;x > 1
    This is ##P(X \leq x)## already, so no more work is needed: ##P(X < 3) = P(X \leq 3) = 1 - 1/9 = 8/9##.

    Incidentally, saying that the function must be a cumulative distribution just because it is denoted by "F" is about the worst justification you could possibly offer. It is a (cumulative) distribution function because it is monotone non-decreasing, and has limits of 0 at -∞ and 1 at +∞. Those are essentially the defining properties of a (cumulative) distribution. Besides that, the problem called it a distribution function, and modern usage leans towards omission of the adjective "cumulative" (so that distribution = cumulative distribution often, nowadays).
  5. Oct 14, 2015 #4
    For the continuous probability distributions, will there ever be a case where P(X<3) is not equivalent to P(X<=3)? or will I always be able to plug it right in like that?
  6. Oct 14, 2015 #5

    Ray Vickson

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    If the (cumulative) distribution function ##F(x)## is continuous, then ##P(X \leq x = P(X < x)## for every ##x##.

    Differences come in when the function ##F(x)## has jump discontinuities; this occurs in so-called "mixed" distributions, which describe random variables that are partially continuous and partly discrete. In such cases, the probability of a single point ##P(X = x)## need not be zero anymore, and we have
    [tex] P(X = x) = P(X \leq x) - P(X < x) = F(x) - F(x-0). [/tex]
    Here, we have adopted the customary convention that ##F(x) = P(X \leq x)## is a right-continuous function (that is, ##\lim_{ y \downarrow x} F(y) = F(x) ##), and ##F(x-0)## is the left-hand limit at ##x##; that is, ##F(x-0) = \lim_{y \uparrow x} F(y)##.

    Such cases are NOT unusual or pathological. They occur, for example, when you truncate a random variable to obtain another one, or when the random variable describes, say, an equipment lifetime that may be ##X = 0## if you bought a "lemon", but is otherwise a continuous random variable in the region ##\{ x > 0 \}##. An example of a truncated random variable might be the lifetime of a piece of equipment that we will scrap for sure if it reaches age = 5 years; otherwise, the lifetime might, for example, be exponential with mean 3 years. In this case, the lifetime ##X## would have a cdf with a jump discontinuity at x = 5:
    [tex] F_X(x) = \begin{cases} 1 - e^{-x/3}, & x < 5 \\
    1, & x \geq 5
    \end{cases} [/tex]
    Here, ##P(X = 5) = e^{-5/3} = F_X(5) - F_X(5-0)##. Other examples abound.
  7. Oct 17, 2015 #6
    Thanks for clearing that up :)
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