Let $(N_t)_{tgeq 0}$ be a Poisson process with parameter $lambda=2$ find $mathbb{E}[N_3N_4].$
$begingroup$
Let $(N_t)_{tgeq 0}$ be a Poisson process with parameter $lambda=2.$
Find $mathbb{E}[N_3N_4].$
The solution here is
begin{align}
mathbb{E}[N_3N_4]&=mathbb{E}[N_3(N_4-N_3+N_3)]tag1\
&=mathbb{E}[N_3(N_4-N_3)+N_3^2]tag2\
&=mathbb{E}[N_3(N_4-N_3)]+mathbb{E}[N_3^2]tag3\
&=mathbb{E}[N_3(N_4-N_3)]+mathbb{Var}[N_3]+mathbb{E}[N_3]^2tag4\
&=mathbb{E}[N_3N_1]+mathbb{Var}[N_3]+mathbb{E}[N_3]^2tag5\
&=mathbb{E}[N_3]mathbb{E}[N_1]+mathbb{Var}[N_3]+mathbb{E}[N_3]^2tag6\
&=6cdot2+6+6^2=54tag7
end{align}
My question is: Why is $N_3$ and $N_1$ independent but not $N_3$ and $N_4$? I assume independence is the reason we can go in (5) from $mathbb{E}[N_3N_1]$ to (6) where we instead get $mathbb{E}[N_3]mathbb{E}[N_1]$?
EDIT/addition:
In the next question in my book, they ask me to explain why the following proof is wrong:
begin{align}
mathbb{E}[(N_3)^2]=mathbb{E}[N_3N_3]=mathbb{E}[N_3(N_6-N_3)]=mathbb{E}[N_3]mathbb{E}[N_6-N_3]=mathbb{E}[N_3]mathbb{E}[N_3]=mathbb{E}[N_3]^2.
end{align}
Their answer is that $N_3neq N_6-N_3$. I find this contradicting because in the previous assignment above, I get to use that $N_4-N_3=N_1$.
probability probability-distributions poisson-process
asked Jan 7 at 16:13
ParsevalParseval
2,7901718
$endgroup$
add a comment |
$begingroup$
Let $(N_t)_{tgeq 0}$ be a Poisson process with parameter $lambda=2.$
Find $mathbb{E}[N_3N_4].$
The solution here is
begin{align}
mathbb{E}[N_3N_4]&=mathbb{E}[N_3(N_4-N_3+N_3)]tag1\
&=mathbb{E}[N_3(N_4-N_3)+N_3^2]tag2\
&=mathbb{E}[N_3(N_4-N_3)]+mathbb{E}[N_3^2]tag3\
&=mathbb{E}[N_3(N_4-N_3)]+mathbb{Var}[N_3]+mathbb{E}[N_3]^2tag4\
&=mathbb{E}[N_3N_1]+mathbb{Var}[N_3]+mathbb{E}[N_3]^2tag5\
&=mathbb{E}[N_3]mathbb{E}[N_1]+mathbb{Var}[N_3]+mathbb{E}[N_3]^2tag6\
&=6cdot2+6+6^2=54tag7
end{align}
My question is: Why is $N_3$ and $N_1$ independent but not $N_3$ and $N_4$? I assume independence is the reason we can go in (5) from $mathbb{E}[N_3N_1]$ to (6) where we instead get $mathbb{E}[N_3]mathbb{E}[N_1]$?
EDIT/addition:
In the next question in my book, they ask me to explain why the following proof is wrong:
begin{align}
mathbb{E}[(N_3)^2]=mathbb{E}[N_3N_3]=mathbb{E}[N_3(N_6-N_3)]=mathbb{E}[N_3]mathbb{E}[N_6-N_3]=mathbb{E}[N_3]mathbb{E}[N_3]=mathbb{E}[N_3]^2.
end{align}
Their answer is that $N_3neq N_6-N_3$. I find this contradicting because in the previous assignment above, I get to use that $N_4-N_3=N_1$.
probability probability-distributions poisson-process
asked Jan 7 at 16:13
ParsevalParseval
2,7901718
$endgroup$
2
$begingroup$
$N_3$ and $N_1$ are not independent, $N_3$ and $N_4-N_3$ is independent due to independent increment and $N_4-N_3$ is equally distributed with $N_1$. There is a minor mistake in the step. It should be $E(N_3(N_4-N_3))=E(N_3)E((N_4-N_3))=E(N_3)E(N_1).$
$endgroup$
– John_Wick
Jan 7 at 16:20
$begingroup$
@John_Wick - This is really confusing, If $N_3$ and $N_4-N_3$ are independent and if $N_4-N_3=N_1$, then $N_3$ and $N_1$ are independent. Same as if $a$ and $b$ are independent and if $b=c+d$ then $a$ and $c+d$ are also independent. What am I missing?
$endgroup$
– Parseval
Jan 7 at 16:26
$begingroup$
"What am I missing?" That $N_4-N_3ne N_1$, not at all, not in the least, absolutely not. But $N_4-N_3stackrel d= N_1$, yes, by definition.
$endgroup$
– Did
Jan 7 at 16:42
$begingroup$
@Did But according to John_Wick's last equations one is led to draw the incorrect conclusion that $N_4-N_3=N_1.$ Guess It's just a matter of notational abuse by my prof, as Michael Lugo points out.
$endgroup$
– Parseval
Jan 7 at 16:48
1
$begingroup$
@Parseval No, John_Wick does not say that $N_4-N_3=N_1$, only that $E(N_4-N_3)=E(N_1)$.
$endgroup$
– Did
Jan 7 at 18:13
add a comment |
$begingroup$
Let $(N_t)_{tgeq 0}$ be a Poisson process with parameter $lambda=2.$
Find $mathbb{E}[N_3N_4].$
The solution here is
begin{align}
mathbb{E}[N_3N_4]&=mathbb{E}[N_3(N_4-N_3+N_3)]tag1\
&=mathbb{E}[N_3(N_4-N_3)+N_3^2]tag2\
&=mathbb{E}[N_3(N_4-N_3)]+mathbb{E}[N_3^2]tag3\
&=mathbb{E}[N_3(N_4-N_3)]+mathbb{Var}[N_3]+mathbb{E}[N_3]^2tag4\
&=mathbb{E}[N_3N_1]+mathbb{Var}[N_3]+mathbb{E}[N_3]^2tag5\
&=mathbb{E}[N_3]mathbb{E}[N_1]+mathbb{Var}[N_3]+mathbb{E}[N_3]^2tag6\
&=6cdot2+6+6^2=54tag7
end{align}
My question is: Why is $N_3$ and $N_1$ independent but not $N_3$ and $N_4$? I assume independence is the reason we can go in (5) from $mathbb{E}[N_3N_1]$ to (6) where we instead get $mathbb{E}[N_3]mathbb{E}[N_1]$?
EDIT/addition:
In the next question in my book, they ask me to explain why the following proof is wrong:
begin{align}
mathbb{E}[(N_3)^2]=mathbb{E}[N_3N_3]=mathbb{E}[N_3(N_6-N_3)]=mathbb{E}[N_3]mathbb{E}[N_6-N_3]=mathbb{E}[N_3]mathbb{E}[N_3]=mathbb{E}[N_3]^2.
end{align}
Their answer is that $N_3neq N_6-N_3$. I find this contradicting because in the previous assignment above, I get to use that $N_4-N_3=N_1$.
probability probability-distributions poisson-process
asked Jan 7 at 16:13
ParsevalParseval
2,7901718
$endgroup$
Let $(N_t)_{tgeq 0}$ be a Poisson process with parameter $lambda=2.$
Find $mathbb{E}[N_3N_4].$
The solution here is
begin{align}
mathbb{E}[N_3N_4]&=mathbb{E}[N_3(N_4-N_3+N_3)]tag1\
&=mathbb{E}[N_3(N_4-N_3)+N_3^2]tag2\
&=mathbb{E}[N_3(N_4-N_3)]+mathbb{E}[N_3^2]tag3\
&=mathbb{E}[N_3(N_4-N_3)]+mathbb{Var}[N_3]+mathbb{E}[N_3]^2tag4\
&=mathbb{E}[N_3N_1]+mathbb{Var}[N_3]+mathbb{E}[N_3]^2tag5\
&=mathbb{E}[N_3]mathbb{E}[N_1]+mathbb{Var}[N_3]+mathbb{E}[N_3]^2tag6\
&=6cdot2+6+6^2=54tag7
end{align}
My question is: Why is $N_3$ and $N_1$ independent but not $N_3$ and $N_4$? I assume independence is the reason we can go in (5) from $mathbb{E}[N_3N_1]$ to (6) where we instead get $mathbb{E}[N_3]mathbb{E}[N_1]$?
EDIT/addition:
In the next question in my book, they ask me to explain why the following proof is wrong:
begin{align}
mathbb{E}[(N_3)^2]=mathbb{E}[N_3N_3]=mathbb{E}[N_3(N_6-N_3)]=mathbb{E}[N_3]mathbb{E}[N_6-N_3]=mathbb{E}[N_3]mathbb{E}[N_3]=mathbb{E}[N_3]^2.
end{align}
Their answer is that $N_3neq N_6-N_3$. I find this contradicting because in the previous assignment above, I get to use that $N_4-N_3=N_1$.
probability probability-distributions poisson-process
probability probability-distributions poisson-process
asked Jan 7 at 16:13
ParsevalParseval
2,7901718
asked Jan 7 at 16:13
ParsevalParseval
2,7901718
edited Jan 7 at 16:30
Parseval
asked Jan 7 at 16:13
ParsevalParseval
2,7901718
asked Jan 7 at 16:13
ParsevalParseval
2,7901718
asked Jan 7 at 16:13
ParsevalParseval
2,7901718
2,7901718
2
$begingroup$
$N_3$ and $N_1$ are not independent, $N_3$ and $N_4-N_3$ is independent due to independent increment and $N_4-N_3$ is equally distributed with $N_1$. There is a minor mistake in the step. It should be $E(N_3(N_4-N_3))=E(N_3)E((N_4-N_3))=E(N_3)E(N_1).$
$endgroup$
– John_Wick
Jan 7 at 16:20
$begingroup$
@John_Wick - This is really confusing, If $N_3$ and $N_4-N_3$ are independent and if $N_4-N_3=N_1$, then $N_3$ and $N_1$ are independent. Same as if $a$ and $b$ are independent and if $b=c+d$ then $a$ and $c+d$ are also independent. What am I missing?
$endgroup$
– Parseval
Jan 7 at 16:26
$begingroup$
"What am I missing?" That $N_4-N_3ne N_1$, not at all, not in the least, absolutely not. But $N_4-N_3stackrel d= N_1$, yes, by definition.
$endgroup$
– Did
Jan 7 at 16:42
$begingroup$
@Did But according to John_Wick's last equations one is led to draw the incorrect conclusion that $N_4-N_3=N_1.$ Guess It's just a matter of notational abuse by my prof, as Michael Lugo points out.
$endgroup$
– Parseval
Jan 7 at 16:48
1
$begingroup$
@Parseval No, John_Wick does not say that $N_4-N_3=N_1$, only that $E(N_4-N_3)=E(N_1)$.
$endgroup$
– Did
Jan 7 at 18:13
add a comment |
2
$begingroup$
$N_3$ and $N_1$ are not independent, $N_3$ and $N_4-N_3$ is independent due to independent increment and $N_4-N_3$ is equally distributed with $N_1$. There is a minor mistake in the step. It should be $E(N_3(N_4-N_3))=E(N_3)E((N_4-N_3))=E(N_3)E(N_1).$
$endgroup$
– John_Wick
Jan 7 at 16:20
$begingroup$
@John_Wick - This is really confusing, If $N_3$ and $N_4-N_3$ are independent and if $N_4-N_3=N_1$, then $N_3$ and $N_1$ are independent. Same as if $a$ and $b$ are independent and if $b=c+d$ then $a$ and $c+d$ are also independent. What am I missing?
$endgroup$
– Parseval
Jan 7 at 16:26
$begingroup$
"What am I missing?" That $N_4-N_3ne N_1$, not at all, not in the least, absolutely not. But $N_4-N_3stackrel d= N_1$, yes, by definition.
$endgroup$
– Did
Jan 7 at 16:42
$begingroup$
@Did But according to John_Wick's last equations one is led to draw the incorrect conclusion that $N_4-N_3=N_1.$ Guess It's just a matter of notational abuse by my prof, as Michael Lugo points out.
$endgroup$
– Parseval
Jan 7 at 16:48
1
$begingroup$
@Parseval No, John_Wick does not say that $N_4-N_3=N_1$, only that $E(N_4-N_3)=E(N_1)$.
$endgroup$
– Did
Jan 7 at 18:13
2
2
$begingroup$
$N_3$ and $N_1$ are not independent, $N_3$ and $N_4-N_3$ is independent due to independent increment and $N_4-N_3$ is equally distributed with $N_1$. There is a minor mistake in the step. It should be $E(N_3(N_4-N_3))=E(N_3)E((N_4-N_3))=E(N_3)E(N_1).$
$endgroup$
– John_Wick
Jan 7 at 16:20
$begingroup$
$N_3$ and $N_1$ are not independent, $N_3$ and $N_4-N_3$ is independent due to independent increment and $N_4-N_3$ is equally distributed with $N_1$. There is a minor mistake in the step. It should be $E(N_3(N_4-N_3))=E(N_3)E((N_4-N_3))=E(N_3)E(N_1).$
$endgroup$
– John_Wick
Jan 7 at 16:20
$begingroup$
@John_Wick - This is really confusing, If $N_3$ and $N_4-N_3$ are independent and if $N_4-N_3=N_1$, then $N_3$ and $N_1$ are independent. Same as if $a$ and $b$ are independent and if $b=c+d$ then $a$ and $c+d$ are also independent. What am I missing?
$endgroup$
– Parseval
Jan 7 at 16:26
$begingroup$
@John_Wick - This is really confusing, If $N_3$ and $N_4-N_3$ are independent and if $N_4-N_3=N_1$, then $N_3$ and $N_1$ are independent. Same as if $a$ and $b$ are independent and if $b=c+d$ then $a$ and $c+d$ are also independent. What am I missing?
$endgroup$
– Parseval
Jan 7 at 16:26
$begingroup$
"What am I missing?" That $N_4-N_3ne N_1$, not at all, not in the least, absolutely not. But $N_4-N_3stackrel d= N_1$, yes, by definition.
$endgroup$
– Did
Jan 7 at 16:42
$begingroup$
"What am I missing?" That $N_4-N_3ne N_1$, not at all, not in the least, absolutely not. But $N_4-N_3stackrel d= N_1$, yes, by definition.
$endgroup$
– Did
Jan 7 at 16:42
$begingroup$
@Did But according to John_Wick's last equations one is led to draw the incorrect conclusion that $N_4-N_3=N_1.$ Guess It's just a matter of notational abuse by my prof, as Michael Lugo points out.
$endgroup$
– Parseval
Jan 7 at 16:48
$begingroup$
@Did But according to John_Wick's last equations one is led to draw the incorrect conclusion that $N_4-N_3=N_1.$ Guess It's just a matter of notational abuse by my prof, as Michael Lugo points out.
$endgroup$
– Parseval
Jan 7 at 16:48
1
1
$begingroup$
@Parseval No, John_Wick does not say that $N_4-N_3=N_1$, only that $E(N_4-N_3)=E(N_1)$.
$endgroup$
– Did
Jan 7 at 18:13
$begingroup$
@Parseval No, John_Wick does not say that $N_4-N_3=N_1$, only that $E(N_4-N_3)=E(N_1)$.
$endgroup$
– Did
Jan 7 at 18:13
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$N_3$ and $N_1$ are not independent. (They can't be! Clearly the time until the third point in your Poisson process depends on the time until the first one - for example, trivially $N_3 > N_1$.) But $N_4 - N_3$ doesn't equal $N_1$, either. This is an abuse of notation.
The $N_1$ in line 5 really should be written as something like $tilde{N}_1$ - it's the first arrival time of the Poisson process starting at time $N_3$. Then you have $E(N_3 (N_4 - N_3)) = E(N_3 tilde{N}_1) = E(N_3) E(tilde{N}_1)$. But since $(tilde{N}_t)_{t ge 0}$ is again a Poisson process with $lambda = 2$ you have $E(tilde{N}_1) = E(N_1)$.
answered Jan 7 at 16:21
Michael LugoMichael Lugo
18k33576
$endgroup$
$begingroup$
Thanks, I see now that the notation confused me. I added something in my question right as you answered, could you please explain if I'm now correct in saying that $$E(N_3)E(N_6-N_3)=E(N_3)E(tilde{N_3})=E(N_3)E(tilde{N_3})neq E(N_3)^2?$$
$endgroup$
– Parseval
Jan 7 at 16:33
$begingroup$
?? Why the last $ne$ sign? Of course $E(N_6-N_3)=E(N_3)$.
$endgroup$
– Did
Jan 8 at 9:53
add a comment |
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$begingroup$
$N_3$ and $N_1$ are not independent. (They can't be! Clearly the time until the third point in your Poisson process depends on the time until the first one - for example, trivially $N_3 > N_1$.) But $N_4 - N_3$ doesn't equal $N_1$, either. This is an abuse of notation.
The $N_1$ in line 5 really should be written as something like $tilde{N}_1$ - it's the first arrival time of the Poisson process starting at time $N_3$. Then you have $E(N_3 (N_4 - N_3)) = E(N_3 tilde{N}_1) = E(N_3) E(tilde{N}_1)$. But since $(tilde{N}_t)_{t ge 0}$ is again a Poisson process with $lambda = 2$ you have $E(tilde{N}_1) = E(N_1)$.
answered Jan 7 at 16:21
Michael LugoMichael Lugo
18k33576
$endgroup$
$begingroup$
Thanks, I see now that the notation confused me. I added something in my question right as you answered, could you please explain if I'm now correct in saying that $$E(N_3)E(N_6-N_3)=E(N_3)E(tilde{N_3})=E(N_3)E(tilde{N_3})neq E(N_3)^2?$$
$endgroup$
– Parseval
Jan 7 at 16:33
$begingroup$
?? Why the last $ne$ sign? Of course $E(N_6-N_3)=E(N_3)$.
$endgroup$
– Did
Jan 8 at 9:53
add a comment |
$begingroup$
$N_3$ and $N_1$ are not independent. (They can't be! Clearly the time until the third point in your Poisson process depends on the time until the first one - for example, trivially $N_3 > N_1$.) But $N_4 - N_3$ doesn't equal $N_1$, either. This is an abuse of notation.
The $N_1$ in line 5 really should be written as something like $tilde{N}_1$ - it's the first arrival time of the Poisson process starting at time $N_3$. Then you have $E(N_3 (N_4 - N_3)) = E(N_3 tilde{N}_1) = E(N_3) E(tilde{N}_1)$. But since $(tilde{N}_t)_{t ge 0}$ is again a Poisson process with $lambda = 2$ you have $E(tilde{N}_1) = E(N_1)$.
answered Jan 7 at 16:21
Michael LugoMichael Lugo
18k33576
$endgroup$
$begingroup$
Thanks, I see now that the notation confused me. I added something in my question right as you answered, could you please explain if I'm now correct in saying that $$E(N_3)E(N_6-N_3)=E(N_3)E(tilde{N_3})=E(N_3)E(tilde{N_3})neq E(N_3)^2?$$
$endgroup$
– Parseval
Jan 7 at 16:33
$begingroup$
?? Why the last $ne$ sign? Of course $E(N_6-N_3)=E(N_3)$.
$endgroup$
– Did
Jan 8 at 9:53
add a comment |
$begingroup$
$N_3$ and $N_1$ are not independent. (They can't be! Clearly the time until the third point in your Poisson process depends on the time until the first one - for example, trivially $N_3 > N_1$.) But $N_4 - N_3$ doesn't equal $N_1$, either. This is an abuse of notation.
The $N_1$ in line 5 really should be written as something like $tilde{N}_1$ - it's the first arrival time of the Poisson process starting at time $N_3$. Then you have $E(N_3 (N_4 - N_3)) = E(N_3 tilde{N}_1) = E(N_3) E(tilde{N}_1)$. But since $(tilde{N}_t)_{t ge 0}$ is again a Poisson process with $lambda = 2$ you have $E(tilde{N}_1) = E(N_1)$.
answered Jan 7 at 16:21
Michael LugoMichael Lugo
18k33576
$endgroup$
$N_3$ and $N_1$ are not independent. (They can't be! Clearly the time until the third point in your Poisson process depends on the time until the first one - for example, trivially $N_3 > N_1$.) But $N_4 - N_3$ doesn't equal $N_1$, either. This is an abuse of notation.
The $N_1$ in line 5 really should be written as something like $tilde{N}_1$ - it's the first arrival time of the Poisson process starting at time $N_3$. Then you have $E(N_3 (N_4 - N_3)) = E(N_3 tilde{N}_1) = E(N_3) E(tilde{N}_1)$. But since $(tilde{N}_t)_{t ge 0}$ is again a Poisson process with $lambda = 2$ you have $E(tilde{N}_1) = E(N_1)$.
answered Jan 7 at 16:21
Michael LugoMichael Lugo
18k33576
edited Jan 7 at 18:25
answered Jan 7 at 16:21
Michael LugoMichael Lugo
18k33576
answered Jan 7 at 16:21
Michael LugoMichael Lugo
18k33576
answered Jan 7 at 16:21
Michael LugoMichael Lugo
18k33576
18k33576
$begingroup$
Thanks, I see now that the notation confused me. I added something in my question right as you answered, could you please explain if I'm now correct in saying that $$E(N_3)E(N_6-N_3)=E(N_3)E(tilde{N_3})=E(N_3)E(tilde{N_3})neq E(N_3)^2?$$
$endgroup$
– Parseval
Jan 7 at 16:33
$begingroup$
?? Why the last $ne$ sign? Of course $E(N_6-N_3)=E(N_3)$.
$endgroup$
– Did
Jan 8 at 9:53
add a comment |
$begingroup$
Thanks, I see now that the notation confused me. I added something in my question right as you answered, could you please explain if I'm now correct in saying that $$E(N_3)E(N_6-N_3)=E(N_3)E(tilde{N_3})=E(N_3)E(tilde{N_3})neq E(N_3)^2?$$
$endgroup$
– Parseval
Jan 7 at 16:33
$begingroup$
?? Why the last $ne$ sign? Of course $E(N_6-N_3)=E(N_3)$.
$endgroup$
– Did
Jan 8 at 9:53
$begingroup$
Thanks, I see now that the notation confused me. I added something in my question right as you answered, could you please explain if I'm now correct in saying that $$E(N_3)E(N_6-N_3)=E(N_3)E(tilde{N_3})=E(N_3)E(tilde{N_3})neq E(N_3)^2?$$
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– Parseval
Jan 7 at 16:33
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Thanks, I see now that the notation confused me. I added something in my question right as you answered, could you please explain if I'm now correct in saying that $$E(N_3)E(N_6-N_3)=E(N_3)E(tilde{N_3})=E(N_3)E(tilde{N_3})neq E(N_3)^2?$$
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– Parseval
Jan 7 at 16:33
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?? Why the last $ne$ sign? Of course $E(N_6-N_3)=E(N_3)$.
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– Did
Jan 8 at 9:53
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?? Why the last $ne$ sign? Of course $E(N_6-N_3)=E(N_3)$.
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– Did
Jan 8 at 9:53
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$N_3$ and $N_1$ are not independent, $N_3$ and $N_4-N_3$ is independent due to independent increment and $N_4-N_3$ is equally distributed with $N_1$. There is a minor mistake in the step. It should be $E(N_3(N_4-N_3))=E(N_3)E((N_4-N_3))=E(N_3)E(N_1).$
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– John_Wick
Jan 7 at 16:20
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@John_Wick - This is really confusing, If $N_3$ and $N_4-N_3$ are independent and if $N_4-N_3=N_1$, then $N_3$ and $N_1$ are independent. Same as if $a$ and $b$ are independent and if $b=c+d$ then $a$ and $c+d$ are also independent. What am I missing?
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– Parseval
Jan 7 at 16:26
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"What am I missing?" That $N_4-N_3ne N_1$, not at all, not in the least, absolutely not. But $N_4-N_3stackrel d= N_1$, yes, by definition.
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– Did
Jan 7 at 16:42
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@Did But according to John_Wick's last equations one is led to draw the incorrect conclusion that $N_4-N_3=N_1.$ Guess It's just a matter of notational abuse by my prof, as Michael Lugo points out.
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– Parseval
Jan 7 at 16:48
1
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@Parseval No, John_Wick does not say that $N_4-N_3=N_1$, only that $E(N_4-N_3)=E(N_1)$.
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– Did
Jan 7 at 18:13