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Evaluating infinite integral for probability density functions

  1. Jul 7, 2010 #1
    1. The problem statement, all variables and given/known data
    So I understand how to evaluate P(4 < X < 6) where the probability density function is f(x)=4x
    I cant seem to understand how to evaluate P(X>6)
    I would have to do something like [tex]\int^6_\infty 4x [/tex] ... so how do I evaluate it?

    2. Relevant equations

    3. The attempt at a solution
    The answer from my book suggests this, after integration = > 2x2 Then
    2([tex]\infty[/tex])2 - 2(6)2
    my 2([tex]\infty[/tex])2 should evaluate to 1.. however in some other problems when its [tex]\int^4_\infty 4x [/tex] (infinity being -infinity) the - [tex]\infty[/tex] part should end up being zero? I'm sorry if this is something basic but there aren't enough examples in my book.
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  3. Jul 7, 2010 #2


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    Remember that a function that is a probability density function must satisfy:

    [tex] f(x) \ge 0 \hbox{ and } \int_{-\infty}^{\infty} f(x)\ dx = 1[/tex]

    f(x) = 4x doesn't satisfy these properties. Frequently pdf's are expressed as piecewise defined functions. For example:

    f(x) = 0 for x ≤ 0, f(x) = 2x for 0 ≤ x ≤ 1, f(x) = 0 for x > 1. Then you would check the property like this:

    [tex]\int_{-\infty}^{\infty} f(x)\ dx = \int_{-\infty}^{0} 0\ dx +\int_{0}^{1} 2x\ dx + \int_{1}^{\infty} 0\ dx[/tex]
  4. Jul 8, 2010 #3
    Im sorry I just used to 4x because when I type the equations in my book the latex just gets messed up (Spent about 20 mins on last post) What I really want to understand is how to evaluate an integral when its infinite.
    I'm going to write integration from a to b as I(a,b) ?

    Printer cartridge has a life modeled by : f(x) = 400x-2 for x >= 400

    Find the probability that such a cartridge has a life of atleast 500 hrs. So P(X >= 500)
    I(infinity, 500) 400x-2 = [-400/x] where x = (infinity, 500)
    The answer is 4/5. How do you arrive at that answer? How do eval the infinity part?

    Really sorry about the representation!!
  5. Jul 8, 2010 #4


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    Those types of integrals are called improper integrals. You do them by using finite limits and let the limits go to infinity:

    [tex]\int_{500}^{\infty} \frac {400}{x^2}\, dx = \lim_{b \rightarrow\infty}\int_{500}^{b} \frac {400}{x^2}\, dx[/tex]

    Just work it out with the b and let b go to infinity in the answer. You will likely find out the term with the b goes to zero since the b is in the denominator.
  6. Jul 9, 2010 #5
    That makes sense.... ok What about situations with a minus infinity and limits that tend to infinity. Like
    [tex]f(x) = \frac{1}{15}(x^2 + 2x)[/tex] find P(X<1.5)
    [tex]\int_{-\infty}^{1.5} f(x) dx[/tex]
    In this case I can see it would tend to zero but what if x were on the denominator? The expression would tend to infinity right?

    Also this function [tex]f(x) =\frac{1}{4}(1 -\frac{x}{8})[/tex] to find P( X > 6)
    [tex]\int_{\infty}^{6} f(x) dx[/tex]
    After integration this would be [tex](\frac{x}{4} - \frac{x^2}{64})[/tex] So again when x = infinity this expression would tend to infinity? Right? The answer suggests it should tend to 1??
    Last edited: Jul 9, 2010
  7. Jul 9, 2010 #6
    Repeat Post!
    Last edited: Jul 9, 2010
  8. Jul 9, 2010 #7


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    You are never going to get anything that makes sense if your f(x) formulas don't represent probability densities. A polynomial can never be a probability density function defined for all x. The only time you will see a polynomial formula for a density is for a piecewise defined function where the polynomial represents f(x) only on a finite region where it is positive and its integral is 1, and f(x) is zero elsewhere. I'm guessing you are missing the "zero elsewhere" part when calculating your integrals.
  9. Jul 10, 2010 #8
    ah, ok I understood quite a bit from that. But there is still just one more problem...

    f(x) =\frac{1}{4}(1 -\frac{x}{8})
    for 0 <= x <= 8, 0 otherwise. Find P(X > 6)

    So after integration [tex]\int_{6}^{\infty }\frac{1}{4}(1 -\frac{x}{8}) dx[/tex] it turns into [tex](\frac{x}{4} - \frac{x^2}{64})[/tex]
    Substituting in x:
    0 - 15/16 = -15/16
    But the answer to this is 1/16 which suggests the infinity part should be one, but it says 0 otherwise right?
  10. Jul 10, 2010 #9


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    Take a look at post #2 in this thread again.
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