Find the probability generating function of $Z=sum_{i=1}^{N}X_i.$
$begingroup$
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
asked Jan 4 at 17:21
ParsevalParseval
3,0001719
$endgroup$
add a comment |
$begingroup$
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
asked Jan 4 at 17:21
ParsevalParseval
3,0001719
$endgroup$
add a comment |
$begingroup$
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
asked Jan 4 at 17:21
ParsevalParseval
3,0001719
$endgroup$
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
3,0001719
asked Jan 4 at 17:21
ParsevalParseval
3,0001719
asked Jan 4 at 17:21
ParsevalParseval
3,0001719
asked Jan 4 at 17:21
ParsevalParseval
3,0001719
asked Jan 4 at 17:21
ParsevalParseval
3,0001719
3,0001719
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Jan 4 at 17:27
angryavianangryavian
42.1k23381
$endgroup$
$begingroup$
Based on what you have written, $s^Z = G_{X_1}(s)...G_{X_N}(s).$ Why is this?
$endgroup$
– Parseval
Jan 4 at 17:30
$begingroup$
@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
$endgroup$
– angryavian
Jan 4 at 17:37
$begingroup$
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.
$endgroup$
– Parseval
Jan 4 at 17:55
1
$begingroup$
@Parseval No, the way I wrote it was a bit misleading, sorry.
$endgroup$
– angryavian
Jan 4 at 18:04
1
$begingroup$
@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.
$endgroup$
– Just_to_Answer
Jan 4 at 20:02
|
show 3 more comments
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1 Answer
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1 Answer
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votes
$begingroup$
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.
answered Jan 4 at 17:27
angryavianangryavian
42.1k23381
$endgroup$
$begingroup$
Based on what you have written, $s^Z = G_{X_1}(s)...G_{X_N}(s).$ Why is this?
$endgroup$
– Parseval
Jan 4 at 17:30
$begingroup$
@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
$endgroup$
– angryavian
Jan 4 at 17:37
$begingroup$
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.
$endgroup$
– Parseval
Jan 4 at 17:55
1
$begingroup$
@Parseval No, the way I wrote it was a bit misleading, sorry.
$endgroup$
– angryavian
Jan 4 at 18:04
1
$begingroup$
@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.
$endgroup$
– Just_to_Answer
Jan 4 at 20:02
|
show 3 more comments
$begingroup$
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.
answered Jan 4 at 17:27
angryavianangryavian
42.1k23381
$endgroup$
$begingroup$
Based on what you have written, $s^Z = G_{X_1}(s)...G_{X_N}(s).$ Why is this?
$endgroup$
– Parseval
Jan 4 at 17:30
$begingroup$
@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
$endgroup$
– angryavian
Jan 4 at 17:37
$begingroup$
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.
$endgroup$
– Parseval
Jan 4 at 17:55
1
$begingroup$
@Parseval No, the way I wrote it was a bit misleading, sorry.
$endgroup$
– angryavian
Jan 4 at 18:04
1
$begingroup$
@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.
$endgroup$
– Just_to_Answer
Jan 4 at 20:02
|
show 3 more comments
$begingroup$
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.
answered Jan 4 at 17:27
angryavianangryavian
42.1k23381
$endgroup$
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.
answered Jan 4 at 17:27
angryavianangryavian
42.1k23381
edited Jan 4 at 18:03
answered Jan 4 at 17:27
angryavianangryavian
42.1k23381
answered Jan 4 at 17:27
angryavianangryavian
42.1k23381
answered Jan 4 at 17:27
angryavianangryavian
42.1k23381
42.1k23381
$begingroup$
Based on what you have written, $s^Z = G_{X_1}(s)...G_{X_N}(s).$ Why is this?
$endgroup$
– Parseval
Jan 4 at 17:30
$begingroup$
@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
$endgroup$
– angryavian
Jan 4 at 17:37
$begingroup$
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.
$endgroup$
– Parseval
Jan 4 at 17:55
1
$begingroup$
@Parseval No, the way I wrote it was a bit misleading, sorry.
$endgroup$
– angryavian
Jan 4 at 18:04
1
$begingroup$
@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.
$endgroup$
– Just_to_Answer
Jan 4 at 20:02
|
show 3 more comments
$begingroup$
Based on what you have written, $s^Z = G_{X_1}(s)...G_{X_N}(s).$ Why is this?
$endgroup$
– Parseval
Jan 4 at 17:30
$begingroup$
@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
$endgroup$
– angryavian
Jan 4 at 17:37
$begingroup$
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.
$endgroup$
– Parseval
Jan 4 at 17:55
1
$begingroup$
@Parseval No, the way I wrote it was a bit misleading, sorry.
$endgroup$
– angryavian
Jan 4 at 18:04
1
$begingroup$
@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.
$endgroup$
– Just_to_Answer
Jan 4 at 20:02
$begingroup$
Based on what you have written, $s^Z = G_{X_1}(s)...G_{X_N}(s).$ Why is this?
$endgroup$
– Parseval
Jan 4 at 17:30
$begingroup$
Based on what you have written, $s^Z = G_{X_1}(s)...G_{X_N}(s).$ Why is this?
$endgroup$
– Parseval
Jan 4 at 17:30
$begingroup$
@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
$endgroup$
– angryavian
Jan 4 at 17:37
$begingroup$
@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
$endgroup$
– angryavian
Jan 4 at 17:37
$begingroup$
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.
$endgroup$
– Parseval
Jan 4 at 17:55
$begingroup$
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.
$endgroup$
– Parseval
Jan 4 at 17:55
1
1
$begingroup$
@Parseval No, the way I wrote it was a bit misleading, sorry.
$endgroup$
– angryavian
Jan 4 at 18:04
$begingroup$
@Parseval No, the way I wrote it was a bit misleading, sorry.
$endgroup$
– angryavian
Jan 4 at 18:04
1
1
$begingroup$
@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.
$endgroup$
– Just_to_Answer
Jan 4 at 20:02
$begingroup$
@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.
$endgroup$
– Just_to_Answer
Jan 4 at 20:02
|
show 3 more comments
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