Conditional probability $P(X_T in B|Y_T, T=t)$

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Let $(X_t)_t geq 0$ and $(Y_t)_tgeq 0$ be stochastic processes where $X_t$ and $Y_t$ are continuous random variables for all $t geq 0$ (with continuous densities). Let $T$ be also a continuous random variable.
Assume that

$P(X_t in B | Y_t) = int_B f(x,Y_t,t), dx $

for all measurable sets $B$ and with a continuous density $f$.

Is it true that $P(X_Tin B | Y_T, T=t) = P(X_t in B| Y_t)$ for $t geq 0$?
Intuitively, it should be true and I would like to prove it using a similar approach like here, i.e. with conditioning on $ t leq T leq t+epsilon $ instead of $T=t$ and then letting $epsilon rightarrow 0$, but I'm having trouble with the dependencies of $X_T$ and $Y_T$ on $T$ and also with how to handle conditioning on $Y_T$ and $T$ at the same time and if a joint density of $X_T, Y_T, T$ or something similar is also needed.

Any help is appreciated! :)

probability probability-theory stochastic-processes stochastic-analysis

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edited Aug 2 at 4:27

asked Aug 1 at 15:03

Max93

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Let $(X_t)_t geq 0$ and $(Y_t)_tgeq 0$ be stochastic processes where $X_t$ and $Y_t$ are continuous random variables for all $t geq 0$ (with continuous densities). Let $T$ be also a continuous random variable.
Assume that

$P(X_t in B | Y_t) = int_B f(x,Y_t,t), dx $

for all measurable sets $B$ and with a continuous density $f$.

Is it true that $P(X_Tin B | Y_T, T=t) = P(X_t in B| Y_t)$ for $t geq 0$?
Intuitively, it should be true and I would like to prove it using a similar approach like here, i.e. with conditioning on $ t leq T leq t+epsilon $ instead of $T=t$ and then letting $epsilon rightarrow 0$, but I'm having trouble with the dependencies of $X_T$ and $Y_T$ on $T$ and also with how to handle conditioning on $Y_T$ and $T$ at the same time and if a joint density of $X_T, Y_T, T$ or something similar is also needed.

Any help is appreciated! :)

probability probability-theory stochastic-processes stochastic-analysis

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edited Aug 2 at 4:27

asked Aug 1 at 15:03

Max93

1838

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

Let $(X_t)_t geq 0$ and $(Y_t)_tgeq 0$ be stochastic processes where $X_t$ and $Y_t$ are continuous random variables for all $t geq 0$ (with continuous densities). Let $T$ be also a continuous random variable.
Assume that

$P(X_t in B | Y_t) = int_B f(x,Y_t,t), dx $

for all measurable sets $B$ and with a continuous density $f$.

Is it true that $P(X_Tin B | Y_T, T=t) = P(X_t in B| Y_t)$ for $t geq 0$?
Intuitively, it should be true and I would like to prove it using a similar approach like here, i.e. with conditioning on $ t leq T leq t+epsilon $ instead of $T=t$ and then letting $epsilon rightarrow 0$, but I'm having trouble with the dependencies of $X_T$ and $Y_T$ on $T$ and also with how to handle conditioning on $Y_T$ and $T$ at the same time and if a joint density of $X_T, Y_T, T$ or something similar is also needed.

Any help is appreciated! :)

probability probability-theory stochastic-processes stochastic-analysis

share|cite|improve this question

edited Aug 2 at 4:27

asked Aug 1 at 15:03

Max93

1838

Let $(X_t)_t geq 0$ and $(Y_t)_tgeq 0$ be stochastic processes where $X_t$ and $Y_t$ are continuous random variables for all $t geq 0$ (with continuous densities). Let $T$ be also a continuous random variable.
Assume that

$P(X_t in B | Y_t) = int_B f(x,Y_t,t), dx $

for all measurable sets $B$ and with a continuous density $f$.

Is it true that $P(X_Tin B | Y_T, T=t) = P(X_t in B| Y_t)$ for $t geq 0$?
Intuitively, it should be true and I would like to prove it using a similar approach like here, i.e. with conditioning on $ t leq T leq t+epsilon $ instead of $T=t$ and then letting $epsilon rightarrow 0$, but I'm having trouble with the dependencies of $X_T$ and $Y_T$ on $T$ and also with how to handle conditioning on $Y_T$ and $T$ at the same time and if a joint density of $X_T, Y_T, T$ or something similar is also needed.

Any help is appreciated! :)

probability probability-theory stochastic-processes stochastic-analysis

share|cite|improve this question

edited Aug 2 at 4:27

asked Aug 1 at 15:03

Max93

1838

share|cite|improve this question

share|cite|improve this question

edited Aug 2 at 4:27

edited Aug 2 at 4:27

edited Aug 2 at 4:27

asked Aug 1 at 15:03

Max93

1838

asked Aug 1 at 15:03

Max93

1838

asked Aug 1 at 15:03

Max93

1838

1838

add a comment | 

add a comment | 

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First, we should define $P(X_Tin B | Y_T, T=t)$. I think, we should assume that $T$ is independent of the processes $(X_t)_t geq 0$ and $(Y_t)_t geq 0$.
Let $g_t$ be the joint density of $X_t, Y_t$,
$$P(X_tin B, Y_tin C) = int_B int_C g_t(x,y) mathrmdy mathrmdx,$$
let $g'_t$ be the marginal density of $Y_t$:
$$P(Y_tin C) = int_C int_Omega g_t(x,y) mathrmdx mathrmdy = int_C g_t'(y) mathrmdy,$$
and let $h$ be the density of $T$:
$$P(T in D) = int_D h(t) mathrmdt $$
The probability distribution of $(X_T, Y_T, T)$ is then
$$P(X_Tin B, Y_Tin C, T in D) = int_D int_B int_C h(t) cdot g_t(x,y) mathrmdy mathrmdx mathrmdt$$
and $i(x,y,t) := h(t)g_t(x,y)$ is the joint density.
Now we can construct the conditional density of $X_T$ given $T=t, Y_T = y$:
$$
j(x|y,t) = fraci(x,y,t)h(t)g_t'(y) = fracg_t(x,y)g'_t(y),
$$
which is by definition a.s. equal to the conditional density of $X_t$ given $Y_t=y$.
We obtain the desired result, when replacing $j(x|y,t)$ with $j(x|Y_t,t)$, which is the conditional density of $X_T$ given $T=t, Y_T$.

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answered Aug 2 at 5:55

Jonas

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1 Answer
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1 Answer
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First, we should define $P(X_Tin B | Y_T, T=t)$. I think, we should assume that $T$ is independent of the processes $(X_t)_t geq 0$ and $(Y_t)_t geq 0$.
Let $g_t$ be the joint density of $X_t, Y_t$,
$$P(X_tin B, Y_tin C) = int_B int_C g_t(x,y) mathrmdy mathrmdx,$$
let $g'_t$ be the marginal density of $Y_t$:
$$P(Y_tin C) = int_C int_Omega g_t(x,y) mathrmdx mathrmdy = int_C g_t'(y) mathrmdy,$$
and let $h$ be the density of $T$:
$$P(T in D) = int_D h(t) mathrmdt $$
The probability distribution of $(X_T, Y_T, T)$ is then
$$P(X_Tin B, Y_Tin C, T in D) = int_D int_B int_C h(t) cdot g_t(x,y) mathrmdy mathrmdx mathrmdt$$
and $i(x,y,t) := h(t)g_t(x,y)$ is the joint density.
Now we can construct the conditional density of $X_T$ given $T=t, Y_T = y$:
$$
j(x|y,t) = fraci(x,y,t)h(t)g_t'(y) = fracg_t(x,y)g'_t(y),
$$
which is by definition a.s. equal to the conditional density of $X_t$ given $Y_t=y$.
We obtain the desired result, when replacing $j(x|y,t)$ with $j(x|Y_t,t)$, which is the conditional density of $X_T$ given $T=t, Y_T$.

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answered Aug 2 at 5:55

Jonas

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up vote
0
down vote

First, we should define $P(X_Tin B | Y_T, T=t)$. I think, we should assume that $T$ is independent of the processes $(X_t)_t geq 0$ and $(Y_t)_t geq 0$.
Let $g_t$ be the joint density of $X_t, Y_t$,
$$P(X_tin B, Y_tin C) = int_B int_C g_t(x,y) mathrmdy mathrmdx,$$
let $g'_t$ be the marginal density of $Y_t$:
$$P(Y_tin C) = int_C int_Omega g_t(x,y) mathrmdx mathrmdy = int_C g_t'(y) mathrmdy,$$
and let $h$ be the density of $T$:
$$P(T in D) = int_D h(t) mathrmdt $$
The probability distribution of $(X_T, Y_T, T)$ is then
$$P(X_Tin B, Y_Tin C, T in D) = int_D int_B int_C h(t) cdot g_t(x,y) mathrmdy mathrmdx mathrmdt$$
and $i(x,y,t) := h(t)g_t(x,y)$ is the joint density.
Now we can construct the conditional density of $X_T$ given $T=t, Y_T = y$:
$$
j(x|y,t) = fraci(x,y,t)h(t)g_t'(y) = fracg_t(x,y)g'_t(y),
$$
which is by definition a.s. equal to the conditional density of $X_t$ given $Y_t=y$.
We obtain the desired result, when replacing $j(x|y,t)$ with $j(x|Y_t,t)$, which is the conditional density of $X_T$ given $T=t, Y_T$.

share|cite|improve this answer

answered Aug 2 at 5:55

Jonas

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up vote
0
down vote

up vote
0
down vote

First, we should define $P(X_Tin B | Y_T, T=t)$. I think, we should assume that $T$ is independent of the processes $(X_t)_t geq 0$ and $(Y_t)_t geq 0$.
Let $g_t$ be the joint density of $X_t, Y_t$,
$$P(X_tin B, Y_tin C) = int_B int_C g_t(x,y) mathrmdy mathrmdx,$$
let $g'_t$ be the marginal density of $Y_t$:
$$P(Y_tin C) = int_C int_Omega g_t(x,y) mathrmdx mathrmdy = int_C g_t'(y) mathrmdy,$$
and let $h$ be the density of $T$:
$$P(T in D) = int_D h(t) mathrmdt $$
The probability distribution of $(X_T, Y_T, T)$ is then
$$P(X_Tin B, Y_Tin C, T in D) = int_D int_B int_C h(t) cdot g_t(x,y) mathrmdy mathrmdx mathrmdt$$
and $i(x,y,t) := h(t)g_t(x,y)$ is the joint density.
Now we can construct the conditional density of $X_T$ given $T=t, Y_T = y$:
$$
j(x|y,t) = fraci(x,y,t)h(t)g_t'(y) = fracg_t(x,y)g'_t(y),
$$
which is by definition a.s. equal to the conditional density of $X_t$ given $Y_t=y$.
We obtain the desired result, when replacing $j(x|y,t)$ with $j(x|Y_t,t)$, which is the conditional density of $X_T$ given $T=t, Y_T$.

share|cite|improve this answer

answered Aug 2 at 5:55

Jonas

259210

First, we should define $P(X_Tin B | Y_T, T=t)$. I think, we should assume that $T$ is independent of the processes $(X_t)_t geq 0$ and $(Y_t)_t geq 0$.
Let $g_t$ be the joint density of $X_t, Y_t$,
$$P(X_tin B, Y_tin C) = int_B int_C g_t(x,y) mathrmdy mathrmdx,$$
let $g'_t$ be the marginal density of $Y_t$:
$$P(Y_tin C) = int_C int_Omega g_t(x,y) mathrmdx mathrmdy = int_C g_t'(y) mathrmdy,$$
and let $h$ be the density of $T$:
$$P(T in D) = int_D h(t) mathrmdt $$
The probability distribution of $(X_T, Y_T, T)$ is then
$$P(X_Tin B, Y_Tin C, T in D) = int_D int_B int_C h(t) cdot g_t(x,y) mathrmdy mathrmdx mathrmdt$$
and $i(x,y,t) := h(t)g_t(x,y)$ is the joint density.
Now we can construct the conditional density of $X_T$ given $T=t, Y_T = y$:
$$
j(x|y,t) = fraci(x,y,t)h(t)g_t'(y) = fracg_t(x,y)g'_t(y),
$$
which is by definition a.s. equal to the conditional density of $X_t$ given $Y_t=y$.
We obtain the desired result, when replacing $j(x|y,t)$ with $j(x|Y_t,t)$, which is the conditional density of $X_T$ given $T=t, Y_T$.

share|cite|improve this answer

answered Aug 2 at 5:55

Jonas

259210

share|cite|improve this answer

share|cite|improve this answer

answered Aug 2 at 5:55

Jonas

259210

answered Aug 2 at 5:55

Jonas

259210

answered Aug 2 at 5:55

Jonas

259210

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add a comment | 

 

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