Poisson-process ${X_t, t ge 0}$ with $lambda>0$. Then $X_t^2$ is a submartingale adapted to the canonical...
Let ${X_t, t ge 0}$ be a homogenous Poisson-process with rate $lambda>0$.
I want to show that both $X_t$ and $X_t^2$ are submartingales adapted to the canonical filtration ${mathcal{F_t}^X, t ge 0}$.
$E[X_t|mathcal{F_s}^X]=E[X_s+(X_t-X_s)|mathcal{F_s}^X]=E[X_s|mathcal{F_s}^X]+E[X_t-X_s|mathcal{F_s}^X]=X_s+E[X_t-X_s]=X_s+(t-s)lambda$
So we have $E[X_t|mathcal{F_s}^X]ge X_s$ which means that $X_t$ is a submartingale adapted to the canonical filtration ${mathcal{F_t}^X, t ge 0}$. Is that correct so far?
I'm struggling to show the same for $X_t^2$ so I could use some help there.
Thanks in advance!
$E[X_t^2|mathcal{F_s}^X]=E[X_s^2+(X_t^2-X_s^2)|mathcal{F_s}^X]=E[X_s^2|mathcal{F_s}^X]+E[X_t^2-X_s^2|mathcal{F_s}^X]=X_s^2+E[X_t^2-X_s^2]ge X_s^2+((t-s)lambda)^2$
probability measure-theory
asked 13 hours ago
user626880
92
add a comment |
Let ${X_t, t ge 0}$ be a homogenous Poisson-process with rate $lambda>0$.
I want to show that both $X_t$ and $X_t^2$ are submartingales adapted to the canonical filtration ${mathcal{F_t}^X, t ge 0}$.
$E[X_t|mathcal{F_s}^X]=E[X_s+(X_t-X_s)|mathcal{F_s}^X]=E[X_s|mathcal{F_s}^X]+E[X_t-X_s|mathcal{F_s}^X]=X_s+E[X_t-X_s]=X_s+(t-s)lambda$
So we have $E[X_t|mathcal{F_s}^X]ge X_s$ which means that $X_t$ is a submartingale adapted to the canonical filtration ${mathcal{F_t}^X, t ge 0}$. Is that correct so far?
I'm struggling to show the same for $X_t^2$ so I could use some help there.
Thanks in advance!
$E[X_t^2|mathcal{F_s}^X]=E[X_s^2+(X_t^2-X_s^2)|mathcal{F_s}^X]=E[X_s^2|mathcal{F_s}^X]+E[X_t^2-X_s^2|mathcal{F_s}^X]=X_s^2+E[X_t^2-X_s^2]ge X_s^2+((t-s)lambda)^2$
probability measure-theory
asked 13 hours ago
user626880
92
Why are you struggling? Either Jensen inequality, or an elementary computations quite similar to the one you present, allow to conclude right away.
– Did
13 hours ago
@Did I tried using Jensen inequality in the last step now, see my edit above. Can I show it like this?
– user626880
13 hours ago
?? You should explain why you think that $$E[X_t^2-X_s^2|mathcal{F_s}^X]=E[X_t^2-X_s^2]$$ and that $$E[X_t^2-X_s^2]ge ((t-s)lambda)^2$$
– Did
12 hours ago
@Did okay I got it now. I also need to determine the Doob decomposition of $X_t$ . Does $M_t=X_t-lambda t$ and $A_t=lambda t$ work?
– user626880
8 hours ago
add a comment |
Let ${X_t, t ge 0}$ be a homogenous Poisson-process with rate $lambda>0$.
I want to show that both $X_t$ and $X_t^2$ are submartingales adapted to the canonical filtration ${mathcal{F_t}^X, t ge 0}$.
$E[X_t|mathcal{F_s}^X]=E[X_s+(X_t-X_s)|mathcal{F_s}^X]=E[X_s|mathcal{F_s}^X]+E[X_t-X_s|mathcal{F_s}^X]=X_s+E[X_t-X_s]=X_s+(t-s)lambda$
So we have $E[X_t|mathcal{F_s}^X]ge X_s$ which means that $X_t$ is a submartingale adapted to the canonical filtration ${mathcal{F_t}^X, t ge 0}$. Is that correct so far?
I'm struggling to show the same for $X_t^2$ so I could use some help there.
Thanks in advance!
$E[X_t^2|mathcal{F_s}^X]=E[X_s^2+(X_t^2-X_s^2)|mathcal{F_s}^X]=E[X_s^2|mathcal{F_s}^X]+E[X_t^2-X_s^2|mathcal{F_s}^X]=X_s^2+E[X_t^2-X_s^2]ge X_s^2+((t-s)lambda)^2$
probability measure-theory
asked 13 hours ago
user626880
92
Let ${X_t, t ge 0}$ be a homogenous Poisson-process with rate $lambda>0$.
I want to show that both $X_t$ and $X_t^2$ are submartingales adapted to the canonical filtration ${mathcal{F_t}^X, t ge 0}$.
$E[X_t|mathcal{F_s}^X]=E[X_s+(X_t-X_s)|mathcal{F_s}^X]=E[X_s|mathcal{F_s}^X]+E[X_t-X_s|mathcal{F_s}^X]=X_s+E[X_t-X_s]=X_s+(t-s)lambda$
So we have $E[X_t|mathcal{F_s}^X]ge X_s$ which means that $X_t$ is a submartingale adapted to the canonical filtration ${mathcal{F_t}^X, t ge 0}$. Is that correct so far?
I'm struggling to show the same for $X_t^2$ so I could use some help there.
Thanks in advance!
$E[X_t^2|mathcal{F_s}^X]=E[X_s^2+(X_t^2-X_s^2)|mathcal{F_s}^X]=E[X_s^2|mathcal{F_s}^X]+E[X_t^2-X_s^2|mathcal{F_s}^X]=X_s^2+E[X_t^2-X_s^2]ge X_s^2+((t-s)lambda)^2$
probability measure-theory
probability measure-theory
asked 13 hours ago
user626880
92
asked 13 hours ago
user626880
92
edited 13 hours ago
asked 13 hours ago
user626880
92
asked 13 hours ago
user626880
92
asked 13 hours ago
user626880
92
92
Why are you struggling? Either Jensen inequality, or an elementary computations quite similar to the one you present, allow to conclude right away.
– Did
13 hours ago
@Did I tried using Jensen inequality in the last step now, see my edit above. Can I show it like this?
– user626880
13 hours ago
?? You should explain why you think that $$E[X_t^2-X_s^2|mathcal{F_s}^X]=E[X_t^2-X_s^2]$$ and that $$E[X_t^2-X_s^2]ge ((t-s)lambda)^2$$
– Did
12 hours ago
@Did okay I got it now. I also need to determine the Doob decomposition of $X_t$ . Does $M_t=X_t-lambda t$ and $A_t=lambda t$ work?
– user626880
8 hours ago
add a comment |
Why are you struggling? Either Jensen inequality, or an elementary computations quite similar to the one you present, allow to conclude right away.
– Did
13 hours ago
@Did I tried using Jensen inequality in the last step now, see my edit above. Can I show it like this?
– user626880
13 hours ago
?? You should explain why you think that $$E[X_t^2-X_s^2|mathcal{F_s}^X]=E[X_t^2-X_s^2]$$ and that $$E[X_t^2-X_s^2]ge ((t-s)lambda)^2$$
– Did
12 hours ago
@Did okay I got it now. I also need to determine the Doob decomposition of $X_t$ . Does $M_t=X_t-lambda t$ and $A_t=lambda t$ work?
– user626880
8 hours ago
Why are you struggling? Either Jensen inequality, or an elementary computations quite similar to the one you present, allow to conclude right away.
– Did
13 hours ago
Why are you struggling? Either Jensen inequality, or an elementary computations quite similar to the one you present, allow to conclude right away.
– Did
13 hours ago
@Did I tried using Jensen inequality in the last step now, see my edit above. Can I show it like this?
– user626880
13 hours ago
@Did I tried using Jensen inequality in the last step now, see my edit above. Can I show it like this?
– user626880
13 hours ago
?? You should explain why you think that $$E[X_t^2-X_s^2|mathcal{F_s}^X]=E[X_t^2-X_s^2]$$ and that $$E[X_t^2-X_s^2]ge ((t-s)lambda)^2$$
– Did
12 hours ago
?? You should explain why you think that $$E[X_t^2-X_s^2|mathcal{F_s}^X]=E[X_t^2-X_s^2]$$ and that $$E[X_t^2-X_s^2]ge ((t-s)lambda)^2$$
– Did
12 hours ago
@Did okay I got it now. I also need to determine the Doob decomposition of $X_t$ . Does $M_t=X_t-lambda t$ and $A_t=lambda t$ work?
– user626880
8 hours ago
@Did okay I got it now. I also need to determine the Doob decomposition of $X_t$ . Does $M_t=X_t-lambda t$ and $A_t=lambda t$ work?
– user626880
8 hours ago
add a comment |
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Why are you struggling? Either Jensen inequality, or an elementary computations quite similar to the one you present, allow to conclude right away.
– Did
13 hours ago
@Did I tried using Jensen inequality in the last step now, see my edit above. Can I show it like this?
– user626880
13 hours ago
?? You should explain why you think that $$E[X_t^2-X_s^2|mathcal{F_s}^X]=E[X_t^2-X_s^2]$$ and that $$E[X_t^2-X_s^2]ge ((t-s)lambda)^2$$
– Did
12 hours ago
@Did okay I got it now. I also need to determine the Doob decomposition of $X_t$ . Does $M_t=X_t-lambda t$ and $A_t=lambda t$ work?
– user626880
8 hours ago