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!

B Finding this probability

  1. Jul 26, 2016 #1
    Hello all,

    I have a question:

    Suppose I want to find the following probability:

    [tex]Pr\left[\frac{\alpha_1}{\alpha_2+1}\leq\gamma\right][/tex]

    where ##\alpha_i## for i=1, 2 is a random variable, whatever the distribution is, and ##\gamma## is a constant. Can I write it as

    [tex]Pr\left[\frac{\alpha_1}{\alpha_2+1}\leq\gamma\right]=\int_{\alpha_2}\Pr\left[\alpha_1\leq\gamma(\alpha_2+1)\right]\,f_{\alpha_2}(\alpha_2)\,d\alpha_2[/tex]

    where ##f_{\alpha_2}(\alpha_2)## is the p.d.f of the random variable ##\alpha_2##, or I need to average over the distribution of ##\gamma(\alpha_2+1)##?

    Thanks
     
  2. jcsd
  3. Jul 26, 2016 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    The integral is exactly this "average". Should work like that, assuming α2+1 > 0.
     
  4. Jul 26, 2016 #3

    mathman

    User Avatar
    Science Advisor
    Gold Member

    [itex]Pr(\frac{\alpha_1}{\alpha_2 +1}\le \gamma)=Pr(\alpha_1 \le (\alpha_2+1)\gamma)=Pr(\alpha_1-\alpha_2 \times \gamma \le \gamma)[/itex].. The last expression is in the form of a sum of (presumed independent) random variables. There is a standard expression, involving convolution of the individual distributions. This assumes [itex]\alpha_2+1 \ge 0[/itex].
     
  5. Jul 26, 2016 #4
    Thanks. Could you give more details on how to find your last probability in terms of the individual pdfs of the independent random variables?
     
  6. Jul 27, 2016 #5

    chiro

    User Avatar
    Science Advisor

    Hey S_David.

    If you have a function of a random variable you can use the transformation theorem to get the new PDF for the new random variable.

    For the sums look at convolution or probability generating functions if they are independent or use the joint and/or conditional distributions to get the probability.
     
  7. Jul 27, 2016 #6

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Just to clarify that: the approach in post 1 is fine, the approach with the convolution will lead to the same thing, just expressed in a different way.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Finding this probability
  1. Probability, The AND (Replies: 19)

Loading...