Find the probability generating function of $Z=sum_{i=1}^{N}X_i.$












0















Let $X_1,X_2,...$ be a sequence of i.i.d Bernoulli random variables
with parameter $p$. Let $N$ be a Poisson random variable with with
parameter $lambda$ which is independet of the $X_i.$



(a) Find the probability generating function of $Z=sum_{i=1}^{N}X_i.$



(b) Use (a) to identify the distribution of $Z$.




The pgf of a sum of i.i.d random variables is the product of their respective pgf's so:



$$G_Z(s)=G_{X_1}(s)cdot G_{X_2}(s)cdot...cdot G_{X_N}(s),$$



and sice the $X_{i}$ are Bernoulli distributed with parameter $p$ we have that



$$G_Z(s)=(1-p+ps)^N.$$



I'm not sure how to proceed from here. I can't express my $G_Z(s)$ in terms of a random variable, so what should I do with the $N$?






probability probability-distributions generating-functions







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    0















    Let $X_1,X_2,...$ be a sequence of i.i.d Bernoulli random variables
    with parameter $p$. Let $N$ be a Poisson random variable with with
    parameter $lambda$ which is independet of the $X_i.$



    (a) Find the probability generating function of $Z=sum_{i=1}^{N}X_i.$



    (b) Use (a) to identify the distribution of $Z$.




    The pgf of a sum of i.i.d random variables is the product of their respective pgf's so:



    $$G_Z(s)=G_{X_1}(s)cdot G_{X_2}(s)cdot...cdot G_{X_N}(s),$$



    and sice the $X_{i}$ are Bernoulli distributed with parameter $p$ we have that



    $$G_Z(s)=(1-p+ps)^N.$$



    I'm not sure how to proceed from here. I can't express my $G_Z(s)$ in terms of a random variable, so what should I do with the $N$?






    probability probability-distributions generating-functions







    share|cite|improve this question

























      0












      0








      0








      Let $X_1,X_2,...$ be a sequence of i.i.d Bernoulli random variables
      with parameter $p$. Let $N$ be a Poisson random variable with with
      parameter $lambda$ which is independet of the $X_i.$



      (a) Find the probability generating function of $Z=sum_{i=1}^{N}X_i.$



      (b) Use (a) to identify the distribution of $Z$.




      The pgf of a sum of i.i.d random variables is the product of their respective pgf's so:



      $$G_Z(s)=G_{X_1}(s)cdot G_{X_2}(s)cdot...cdot G_{X_N}(s),$$



      and sice the $X_{i}$ are Bernoulli distributed with parameter $p$ we have that



      $$G_Z(s)=(1-p+ps)^N.$$



      I'm not sure how to proceed from here. I can't express my $G_Z(s)$ in terms of a random variable, so what should I do with the $N$?






      probability probability-distributions generating-functions







      share|cite|improve this question














      Let $X_1,X_2,...$ be a sequence of i.i.d Bernoulli random variables
      with parameter $p$. Let $N$ be a Poisson random variable with with
      parameter $lambda$ which is independet of the $X_i.$



      (a) Find the probability generating function of $Z=sum_{i=1}^{N}X_i.$



      (b) Use (a) to identify the distribution of $Z$.




      The pgf of a sum of i.i.d random variables is the product of their respective pgf's so:



      $$G_Z(s)=G_{X_1}(s)cdot G_{X_2}(s)cdot...cdot G_{X_N}(s),$$



      and sice the $X_{i}$ are Bernoulli distributed with parameter $p$ we have that



      $$G_Z(s)=(1-p+ps)^N.$$



      I'm not sure how to proceed from here. I can't express my $G_Z(s)$ in terms of a random variable, so what should I do with the $N$?






      probability probability-distributions generating-functions




      probability probability-distributions generating-functions






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      asked Jan 4 at 17:21









      ParsevalParseval

      2,7671718




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          1 Answer
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          Long story short, it ends up being the expectation of what you have written. But you should start from scratch and try to use the tower property by conditioning on $N$ to formalize all this.




          $$G_Z(s) = E[s^Z] = E[E[s^Z mid N]] = E[E[s^{X_1} cdots s^{X_N} mid N]] = E[(1-p+ps)^N]= e^{lambda p(s-1)}.$$ It remains to identify the final PGF.







          share|cite|improve this answer























          • Based on what you have written, $s^Z = G_{X_1}(s)...G_{X_N}(s).$ Why is this?
            – Parseval
            Jan 4 at 17:30










          • @Parseval conditioned on $N$ (loosely, "when $N$ is fixed"), you may use the "PGF of sum of independent RVs is product of PGFs" result that you stated
            – angryavian
            Jan 4 at 17:37










          • Yes, but according to the book, that proposition is $$G_Z(s)=G_{X_1}(s)...G_{X_n}(s)$$ and not $$s^Z=G_{X_1}(s)...G_{X_n}(s).$$ I'm probably misunderstanding something.
            – Parseval
            Jan 4 at 17:55








          • 1




            @Parseval No, the way I wrote it was a bit misleading, sorry.
            – angryavian
            Jan 4 at 18:04






          • 1




            @Parseval: Once you take out $e^{-lambda}$, what remains is a power series of $e^x$ where $x=(1-p+ps)lambda$. Then it all simplifies to what you want.
            – Just_to_Answer
            Jan 4 at 20:02











          Your Answer





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          1 Answer
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          1 Answer
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          active

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          active

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          2














          Long story short, it ends up being the expectation of what you have written. But you should start from scratch and try to use the tower property by conditioning on $N$ to formalize all this.




          $$G_Z(s) = E[s^Z] = E[E[s^Z mid N]] = E[E[s^{X_1} cdots s^{X_N} mid N]] = E[(1-p+ps)^N]= e^{lambda p(s-1)}.$$ It remains to identify the final PGF.







          share|cite|improve this answer























          • Based on what you have written, $s^Z = G_{X_1}(s)...G_{X_N}(s).$ Why is this?
            – Parseval
            Jan 4 at 17:30










          • @Parseval conditioned on $N$ (loosely, "when $N$ is fixed"), you may use the "PGF of sum of independent RVs is product of PGFs" result that you stated
            – angryavian
            Jan 4 at 17:37










          • Yes, but according to the book, that proposition is $$G_Z(s)=G_{X_1}(s)...G_{X_n}(s)$$ and not $$s^Z=G_{X_1}(s)...G_{X_n}(s).$$ I'm probably misunderstanding something.
            – Parseval
            Jan 4 at 17:55








          • 1




            @Parseval No, the way I wrote it was a bit misleading, sorry.
            – angryavian
            Jan 4 at 18:04






          • 1




            @Parseval: Once you take out $e^{-lambda}$, what remains is a power series of $e^x$ where $x=(1-p+ps)lambda$. Then it all simplifies to what you want.
            – Just_to_Answer
            Jan 4 at 20:02
















          2














          Long story short, it ends up being the expectation of what you have written. But you should start from scratch and try to use the tower property by conditioning on $N$ to formalize all this.




          $$G_Z(s) = E[s^Z] = E[E[s^Z mid N]] = E[E[s^{X_1} cdots s^{X_N} mid N]] = E[(1-p+ps)^N]= e^{lambda p(s-1)}.$$ It remains to identify the final PGF.







          share|cite|improve this answer























          • Based on what you have written, $s^Z = G_{X_1}(s)...G_{X_N}(s).$ Why is this?
            – Parseval
            Jan 4 at 17:30










          • @Parseval conditioned on $N$ (loosely, "when $N$ is fixed"), you may use the "PGF of sum of independent RVs is product of PGFs" result that you stated
            – angryavian
            Jan 4 at 17:37










          • Yes, but according to the book, that proposition is $$G_Z(s)=G_{X_1}(s)...G_{X_n}(s)$$ and not $$s^Z=G_{X_1}(s)...G_{X_n}(s).$$ I'm probably misunderstanding something.
            – Parseval
            Jan 4 at 17:55








          • 1




            @Parseval No, the way I wrote it was a bit misleading, sorry.
            – angryavian
            Jan 4 at 18:04






          • 1




            @Parseval: Once you take out $e^{-lambda}$, what remains is a power series of $e^x$ where $x=(1-p+ps)lambda$. Then it all simplifies to what you want.
            – Just_to_Answer
            Jan 4 at 20:02














          2












          2








          2






          Long story short, it ends up being the expectation of what you have written. But you should start from scratch and try to use the tower property by conditioning on $N$ to formalize all this.




          $$G_Z(s) = E[s^Z] = E[E[s^Z mid N]] = E[E[s^{X_1} cdots s^{X_N} mid N]] = E[(1-p+ps)^N]= e^{lambda p(s-1)}.$$ It remains to identify the final PGF.







          share|cite|improve this answer














          Long story short, it ends up being the expectation of what you have written. But you should start from scratch and try to use the tower property by conditioning on $N$ to formalize all this.




          $$G_Z(s) = E[s^Z] = E[E[s^Z mid N]] = E[E[s^{X_1} cdots s^{X_N} mid N]] = E[(1-p+ps)^N]= e^{lambda p(s-1)}.$$ It remains to identify the final PGF.








          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 4 at 18:03

























          answered Jan 4 at 17:27









          angryavianangryavian

          39.3k23280




          39.3k23280












          • Based on what you have written, $s^Z = G_{X_1}(s)...G_{X_N}(s).$ Why is this?
            – Parseval
            Jan 4 at 17:30










          • @Parseval conditioned on $N$ (loosely, "when $N$ is fixed"), you may use the "PGF of sum of independent RVs is product of PGFs" result that you stated
            – angryavian
            Jan 4 at 17:37










          • Yes, but according to the book, that proposition is $$G_Z(s)=G_{X_1}(s)...G_{X_n}(s)$$ and not $$s^Z=G_{X_1}(s)...G_{X_n}(s).$$ I'm probably misunderstanding something.
            – Parseval
            Jan 4 at 17:55








          • 1




            @Parseval No, the way I wrote it was a bit misleading, sorry.
            – angryavian
            Jan 4 at 18:04






          • 1




            @Parseval: Once you take out $e^{-lambda}$, what remains is a power series of $e^x$ where $x=(1-p+ps)lambda$. Then it all simplifies to what you want.
            – Just_to_Answer
            Jan 4 at 20:02


















          • Based on what you have written, $s^Z = G_{X_1}(s)...G_{X_N}(s).$ Why is this?
            – Parseval
            Jan 4 at 17:30










          • @Parseval conditioned on $N$ (loosely, "when $N$ is fixed"), you may use the "PGF of sum of independent RVs is product of PGFs" result that you stated
            – angryavian
            Jan 4 at 17:37










          • Yes, but according to the book, that proposition is $$G_Z(s)=G_{X_1}(s)...G_{X_n}(s)$$ and not $$s^Z=G_{X_1}(s)...G_{X_n}(s).$$ I'm probably misunderstanding something.
            – Parseval
            Jan 4 at 17:55








          • 1




            @Parseval No, the way I wrote it was a bit misleading, sorry.
            – angryavian
            Jan 4 at 18:04






          • 1




            @Parseval: Once you take out $e^{-lambda}$, what remains is a power series of $e^x$ where $x=(1-p+ps)lambda$. Then it all simplifies to what you want.
            – Just_to_Answer
            Jan 4 at 20:02
















          Based on what you have written, $s^Z = G_{X_1}(s)...G_{X_N}(s).$ Why is this?
          – Parseval
          Jan 4 at 17:30




          Based on what you have written, $s^Z = G_{X_1}(s)...G_{X_N}(s).$ Why is this?
          – Parseval
          Jan 4 at 17:30












          @Parseval conditioned on $N$ (loosely, "when $N$ is fixed"), you may use the "PGF of sum of independent RVs is product of PGFs" result that you stated
          – angryavian
          Jan 4 at 17:37




          @Parseval conditioned on $N$ (loosely, "when $N$ is fixed"), you may use the "PGF of sum of independent RVs is product of PGFs" result that you stated
          – angryavian
          Jan 4 at 17:37












          Yes, but according to the book, that proposition is $$G_Z(s)=G_{X_1}(s)...G_{X_n}(s)$$ and not $$s^Z=G_{X_1}(s)...G_{X_n}(s).$$ I'm probably misunderstanding something.
          – Parseval
          Jan 4 at 17:55






          Yes, but according to the book, that proposition is $$G_Z(s)=G_{X_1}(s)...G_{X_n}(s)$$ and not $$s^Z=G_{X_1}(s)...G_{X_n}(s).$$ I'm probably misunderstanding something.
          – Parseval
          Jan 4 at 17:55






          1




          1




          @Parseval No, the way I wrote it was a bit misleading, sorry.
          – angryavian
          Jan 4 at 18:04




          @Parseval No, the way I wrote it was a bit misleading, sorry.
          – angryavian
          Jan 4 at 18:04




          1




          1




          @Parseval: Once you take out $e^{-lambda}$, what remains is a power series of $e^x$ where $x=(1-p+ps)lambda$. Then it all simplifies to what you want.
          – Just_to_Answer
          Jan 4 at 20:02




          @Parseval: Once you take out $e^{-lambda}$, what remains is a power series of $e^x$ where $x=(1-p+ps)lambda$. Then it all simplifies to what you want.
          – Just_to_Answer
          Jan 4 at 20:02


















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