Find the probability generating function of $Z=sum_{i=1}^{N}X_i.$
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
add a comment |
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
add a comment |
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
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
asked Jan 4 at 17:21
ParsevalParseval
2,7671718
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.
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
|
show 3 more comments
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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.
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
|
show 3 more comments
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.
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
|
show 3 more comments
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.
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.
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
|
show 3 more comments
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
|
show 3 more comments
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