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CDF and MGF Relation

  1. Oct 29, 2009 #1
    Hello,

    Suppose that the Cumulative Distribution Function (CDF) of a random variable X is [tex]F_X(x)[/tex], which is by definition is:

    [tex]F_X(x)=\text{Pr}\left[X\leq x\right]=\text{Pr}\left[\frac{1}{X}\geq \frac{1}{x}\right]=1-\text{Pr}\left[\frac{1}{X}\leq \frac{1}{x}\right]=1-F_{1/X}\left(1/x\right)[/tex]

    Considering this relation between the CDF of X and the CDF of its reciprocal, what is the relation between the Moment Generating Function (MGF) of X and its reciprocal?

    Any help will be highly appreciated.

    Thanks in advance
     
  2. jcsd
  3. Oct 31, 2009 #2
    A good starting point would be to think of a relation between the CDF of X and the MGF of X, wouldn't it?
     
  4. Oct 31, 2009 #3
    Yes right, and I know what is the relation between them, but I want to see if another one has another idea. Anyway, the relation is:

    [tex]M_X(s)=s\mathcal{L}\left\{F_X(x)\right\}[/tex]

    I have tried this, and it yields no where.

    Regards
     
  5. Oct 31, 2009 #4
    What would the CDF and MGF look like if X is uniform on [0,1] ?
     
  6. Nov 27, 2009 #5
    The CDF of a uniformly distributed random variable X is:

    [tex]F_X(x)=\begin{cases}0&x<0\\\frac{x-a}{b-a}&a\leq x<b\\1&x\ge b\end{cases}[/tex]

    Here, it may easier to derive the MGF from the PDF, not from the CDF. The PDF of X will be:

    [tex]f_X(x)=\begin{cases}\frac{1}{b-a}&a\leq x\leq b\\0&\mbox{elsewhere}\end{cases}[/tex]

    Then the MGF of X is:

    [tex]\mathcal{M}_X(s)=E_X\left[\text{e}^{sx}\right]=\int_a^b\text{e}^{sx}f_X(x)\,dx=\frac{\text{e}^{bs}-\text{e}^{as}}{s\left(b-a\right)}[/tex]

    But, what is the relation of this to the primary question?

    Anyway, I have found the following relations between the MGF of X and the MGF of its reciprocal:

    [tex]
    \mathcal{M}_X(s)=\int_s^{\infty}\int_0^{\infty}J_0\left(2\,\sqrt{u\,p}\right)\mathcal{M}_{1/X}(s)\,du\,dp\\
    [/tex]
    [tex]\mathcal{M}_{1/X}(s)=1-\sqrt{s}\int_0^{\pi/2}\frac{\sec^2 (\zeta)}{\sqrt{\tan (\zeta)}}J_1\left(2\,\sqrt{s\tan (\zeta)}\right)\mathcal{M}_X\left(\tan (\zeta)\right)\,d\zeta
    [/tex]

    where [tex]J_v(.)[/tex] is the vth order bessel function of the first kind. I do not know how they got there. Does anybody know how to derive these relations?

    Thanks in advance
     
  7. Nov 27, 2009 #6

    mathman

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    Gold Member

    If X < 0 and x > 0, your statement about reciprocals doesn't hold.
     
  8. Nov 28, 2009 #7
    Yes , I forgot to mention that [tex]0\leq X, x<\infty[/tex]. Then, is there any problem?

    Regards
     
  9. Nov 28, 2009 #8

    mathman

    User Avatar
    Science Advisor
    Gold Member

    Not in your original statement.
     
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