+ Reply to Thread
Results 1 to 3 of 3

Thread: Symmetry of Cumulative St. Normal Function

  1. #1
    Points: 740, Level: 14
    Level completed: 40%, Points required for next Level: 60

    Posts
    31
    Thanks
    11
    Thanked 1 Time in 1 Post

    Symmetry of Cumulative St. Normal Function




    Hi all,

    I have to show that for the normal distribution F(z)=1-F(-z), starting with the integral:

    \displaystyle F(z)=\int_{-\infty} ^z \frac{1}{\sqrt{2\pi}}e^{-w^2/2}dw

    Now with a small change of variables w=-v, unless I am wrong, the integral becomes:

    \displaystyle -\int_{-z}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-v^2/2}dv which is not what it's supposed to be because the minus in front, which comes from the Jacobian, screws things up. Have I done something wrong?
    Thanks.

  2. #2
    TS Contributor
    Points: 22,410, Level: 93
    Level completed: 6%, Points required for next Level: 940

    Posts
    3,020
    Thanks
    12
    Thanked 565 Times in 537 Posts

    Re: Symmetry of Cumulative St. Normal Function

    you should obtain

    -\int_{\infty}^{-z} f(v)dv = \int_{-z}^{\infty} f(v)dv

    It is because you let w = -v.

    When w = -\infty, v = +\infty; w = z, v = -z; dw = -dv; w^2 = v^2

    And with the above result the claim becomes obvious.

  3. The Following User Says Thank You to BGM For This Useful Post:

    JohnK (10-07-2013)

  4. #3
    Points: 740, Level: 14
    Level completed: 40%, Points required for next Level: 60

    Posts
    31
    Thanks
    11
    Thanked 1 Time in 1 Post

    Re: Symmetry of Cumulative St. Normal Function


    Thanks, I mistakingly switched the limits of integration placing the infinity limit the upper one, but I see now that what I did was incorrect.

+ Reply to Thread

           




Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts






Advertise on Talk Stats