1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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

    User Avatar
    Science Advisor
    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

    User Avatar
    Science Advisor
    Homework Helper

    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 :)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted