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Finding A. Prob.

  1. Oct 22, 2013 #1
    1. The problem statement, all variables and given/known data

    The expoential distribution is given by

    f(x) = 0 for x <0 and f(x) = Ae^-cx for x >= 0

    2. Relevant equations



    3. The attempt at a solution

    Find A so that f(x) is a probability distribution. So all that I did was use the def.

    I said A∫e^(-cx) dx = 1 from 0 to ∞ I integrated it I got that A/c = 1 so A = c.


    So then the next question says find P(x <1) so I did ∫ce^-ct dt = -e^(-ct) between [0,1] so I'm just confused because I have c still it isn't like it's going to drop out. So I'm thinking I did something wrong. Should A be numerical? I don't see how . Thanks
     
  2. jcsd
  3. Oct 22, 2013 #2
    Probably a statistician would know this, but I don't. What is P(x)?
     
  4. Oct 22, 2013 #3
    Probability. Idk. If they say find the probability that x <1 then you express it as P( x<1)
     
  5. Oct 22, 2013 #4

    Mark44

    Staff: Mentor

    Since you aren't given a value for c, you aren't going to get a value for A that is a specific number. I don't see anything wrong in your work.
     
  6. Oct 22, 2013 #5
    OK. So report my answer for P( x<1 ) = ∫ce^-ct dt = -e^(-ct) between [0,1]
    as 1/e^c
     
  7. Oct 22, 2013 #6

    Mark44

    Staff: Mentor

    I'm not sure you're writing what you mean.

    $$ \int e^{-ct}dt = -\frac 1 c \int e^{-ct} (-c dt) = -\frac 1 c e^{-ct} + C$$

    For the definite integral you get 1/c. Note that I haven't included A, but if you multiply throughout by A, you get A/c. For that to be equal to 1, then A = c, which is what you had.
     
  8. Oct 22, 2013 #7

    LCKurtz

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    That isn't correct. ##P(X<1)=\int_0^1 ce^{-cx}~dx## and you have left out the lower limit in your calculation.

    Also, I think most texts define the exponential density function as ##\lambda e^{-\lambda x}## for ##x > 0## in the first place.
     
  9. Oct 22, 2013 #8
    This was my course pack. What do you mean it isn't right ? What isn't right? my answer for my integral 1/e^c?

    I have the lower limit [0,1] what are you talking about.
     
  10. Oct 22, 2013 #9
    OH I see the lower limit missing. But what else isn't right?
     
  11. Oct 22, 2013 #10
    ## P(X<1)=\int_0^1 ce^{-cx}~dx = 1-1/e^c ## Is this correct?

    And is
    ## P(.5 < X<1.5)=\int_0^1 ce^{-cx}~dx = 1/e^c(0.5) - 1/e^c(1.5) ## correct?
    I don't get what your saying is wrong LCKurtz
     
    Last edited: Oct 22, 2013
  12. Oct 22, 2013 #11

    Mark44

    Staff: Mentor

    I think so. It's been a long time since I studied mathematical statistics, and I seem to recall that this is a Poisson distribution.
    ## P(.5 < X<1.5)=\int_{.5}^{1.5} ce^{-cx}~dx ##

    In general, ## P(a < X< b)=\int_a^b ce^{-cx}~dx ##, where a ≥ 0 and b ≥ a.
     
  13. Oct 22, 2013 #12

    LCKurtz

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    Yes

    Put the correct limits on the integral and it will be.
     
  14. Oct 22, 2013 #13
    Oh I'm sorry. I just copied your tex and forgot to change it. It is OK on my paper.
    Thanks
     
  15. Oct 22, 2013 #14

    Ray Vickson

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    The value A = c is correct. Also:
    [tex]P(X \leq v) = 1 - e^{-cv}, \: \: P(X > v) = e^{-cv}.[/tex]
    These are used so often that if you plan to do more Probability in the future you should commit them to memory, along with
    [tex] E X = \frac{1}{c}, \; \; \text{Var} X = \frac{1}{c^2}. [/tex]

    BTW: the standard way of writing is to use ##e^{-cv}## rather than ##1/e^{cv}##. They are, of course, eqiuivalent, but the first way is used all the time by everybody, at least in works on probability.
     
    Last edited: Oct 22, 2013
  16. Oct 22, 2013 #15
    OH Ok cool thanks for the info ray.
     
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