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Math Help - Probability Generating Function.

  1. #1
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    Probability Generating Function.

    I found this in one of the review exercises.

    Let X be a discrete random variable with values in \mathbb Z_{+} := {0,1,2, ...} with probability generating function

    P_{X}(z) = \mathbb E(z^{X}) = \sum^{+ \infty}_{k=0} \mathbb P (X = k)z^{k}

    (i) Show that P_X defines a mapping from [0,1] to [0,1].
    (ii) Let (X_n) be a sequence of i.i.d random variables with values in \mathbb Z_{+} and T a random variable with values in \mathbb Z_{+} which is independent of (X_n). Define

    S_{0} = 0 and S_{T} = \sum^{T}_{j=1} X_{j}

    Show that if we denote by P_{T} and P_{X} the probability the generating functions of T and X_{1} respectively, then the probability generating function of S is given by

    \forall z \in [0,1], P_{S}(z) = P_{T}(P_{X}(z)).

    Thank you (:
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  2. #2
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    Hello,
    Quote Originally Posted by panda* View Post
    I found this in one of the review exercises.

    Let X be a discrete random variable with values in \mathbb Z_{+} := {0,1,2, ...} with probability generating function

    P_{X}(z) = \mathbb E(z^{X}) = \sum^{+ \infty}_{k=0} \mathbb P (X = k)z^{k}

    (i) Show that P_X defines a mapping from [0,1] to [0,1].
    I'll think on this one later ^^
    (ii) Let (X_n) be a sequence of i.i.d random variables with values in \mathbb Z_{+} and T a random variable with values in \mathbb Z_{+} which is independent of (X_n). Define

    S_{0} = 0 and S_{T} = \sum^{T}_{j=1} X_{j}

    Show that if we denote by P_{T} and P_{X} the probability the generating functions of T and X_{1} respectively, then the probability generating function of S is given by

    \forall z \in [0,1], P_{S}(z) = P_{T}(P_{X}(z)).

    Thank you (:
    Like for any problem of this kind, take a partition of the set of events.
    Namely \Omega=\bigcup_{k=0}^\infty \{T=k\}

    ({.} denotes an event... if you prefer, it's the set of \omega\in\Omega such that the even happens)

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

    Thus
    \begin{aligned}<br />
\{S_T=n\}&=\{S_T=n\}\cap \Omega \\<br />
&=\{S_T=n\}\cap \bigcup_{k=0}^\infty \{T=k\} \\<br />
&=\bigcup_{k=0}^\infty \{S_T=n ~,~ T=k\} \quad (\star)<br />
\end{aligned}

    Hence :
    \begin{aligned}<br />
\mathbb{P}(S_T=n) &=\mathbb{P}\left(\bigcup_{k=0}^\infty \{S_T=n ~,~ T=k\}\right) \\<br />
&=\sum_{k=0}^\infty \mathbb{P}(S_T=n ~,~ T=k) \quad (\star\star) \\<br />
&=\sum_{k=0}^\infty \mathbb{P}(S_T=n \mid T=k)\mathbb{P}(T=k) \quad (\star\star\star) \\<br />
&=\sum_{k=0}^\infty \mathbb{P}(S_k=n)\mathbb{P}(T=k) \end{aligned}

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

    We have :
    \begin{aligned}P_S(z) &=\mathbb{E}(z^{S_T}) \\<br />
&=\sum_{n=0}^\infty \mathbb{P}(S_T=n)z^n \\<br />
&=\sum_{n=0}^\infty \sum_{k=0}^\infty \mathbb{P}(S_k=n)\mathbb{P}(T=k)z^n \\<br />
&=\sum_{k=0}^\infty \mathbb{P}(T=k) \sum_{n=0}^\infty \mathbb{P}(S_k=n)z^n \end{aligned}
    {\color{white}P_S(z)}=\sum_{k=0}^\infty \mathbb{P}(T=k) P_{S_k}(z)


    But what is P_{S_k}(z) ?
    S_k=\sum_{j=1}^k

    Hence P_{S_k}(z)=\mathbb{E}(z^{S_k})=\mathbb{E}\left(z^{  X_1+\dots+X_k}\right)=\mathbb{E}(z^{X_1}\cdots z^{X_k})

    Since the Xi's are independent, this is equal to the product of the expectations :
    P_{S_k}(z)=\mathbb{E}(z^{X_1})\cdots\mathbb{E}(z^{  X_k})

    Since the Xi's follow the same distribution as X, this is :
    P_{S_k}(z)=[P_X(z)]^k


    Now finally, we have :
    P_S(z)=\sum_{k=0}^\infty \mathbb{P}(T=k)[P_X(z)]^k

    Can you recognize that it's the pgf of T, taken at a particular value ?
    Spoiler:
    namely at P_X(z)


    And you're done... I hope you like it, because it's been quite loooooooong to type lol!

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

    (\star) is obtained just by using the distributive law of intersection and union. It shouldn't be difficult to prove it.

    (\star\star) because for k\neq m, {T=k} and {T=m} are disjoint (intersection = empty set). And hence { S_T=n , T=k} and { S_T=n , T=m} are disjoint.
    The equality follows from the third axiom of Kolmogorov.

    (\star\star\star) is actually Bayes' formula





    Edit : lol...that's my second 5th post XD
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  3. #3
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    Quote Originally Posted by panda* View Post
    I found this in one of the review exercises.

    Let X be a discrete random variable with values in \mathbb Z_{+} := {0,1,2, ...} with probability generating function

    P_{X}(z) = \mathbb E(z^{X}) = \sum^{+ \infty}_{k=0} \mathbb P (X = k)z^{k}

    (i) Show that P_X defines a mapping from [0,1] to [0,1].
    [snip]
    Observe that P_X(0) = P(X = 0), P_X(1) = 1, and P_X(z) is an increasing function of z.
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  4. #4
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    Moo, thanks so much for your effort! I understand better now. Thank you again.

    Awkward, is that all you have to show?
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