Obtaining derivatives for the moments of random variables
$begingroup$
I am currently working through a problem and am having trouble interpreting my results.
I start with a system whose dynamics are described by the equation:
$$
frac{dC}{dt} = f(C(t),theta)
$$
where $theta$ is a random variable independent of time meaning $C(t)$ too becomes a random variable. We can write the equation for the change in the first moment $E[C(t)]$ as:
$$
frac{dE[C(t)]}{dt} = frac{d}{dt}int p_C(C(t)) C(t) dC
$$
where $p_C(C)$ is the probability distribution function for C. Recalling that
$$
E[g(x)] = int g(x) p_x(x) dx
$$
and rearanging the first equation
$$
g(theta) = C(t) = int f(C(t),theta) dt
$$
we can combine to get
begin{align}
frac{dE[C(t)]}{dt} &= frac{d}{dt}int p_{theta}(theta) g(theta) dtheta \
&= frac{d}{dt}int p_{theta}(theta) int f(C(t),theta) dt dtheta \
&= int p_{theta}(theta)f(C(t),theta) dtheta \
&= E[f(C(t),theta)]
end{align}
I am wondering
- Is the working correct?
- How to interpret the final result. This expectation does not include the variance in $C(t)$ but does this matter? What does this mean if we take a simple form (i.e. $f(C(t),theta) = theta C(t) - C(t)^2$)?
ordinary-differential-equations probability-theory expected-value sde
asked Jan 7 at 22:25
TomTom
436
$endgroup$
add a comment |
$begingroup$
I am currently working through a problem and am having trouble interpreting my results.
I start with a system whose dynamics are described by the equation:
$$
frac{dC}{dt} = f(C(t),theta)
$$
where $theta$ is a random variable independent of time meaning $C(t)$ too becomes a random variable. We can write the equation for the change in the first moment $E[C(t)]$ as:
$$
frac{dE[C(t)]}{dt} = frac{d}{dt}int p_C(C(t)) C(t) dC
$$
where $p_C(C)$ is the probability distribution function for C. Recalling that
$$
E[g(x)] = int g(x) p_x(x) dx
$$
and rearanging the first equation
$$
g(theta) = C(t) = int f(C(t),theta) dt
$$
we can combine to get
begin{align}
frac{dE[C(t)]}{dt} &= frac{d}{dt}int p_{theta}(theta) g(theta) dtheta \
&= frac{d}{dt}int p_{theta}(theta) int f(C(t),theta) dt dtheta \
&= int p_{theta}(theta)f(C(t),theta) dtheta \
&= E[f(C(t),theta)]
end{align}
I am wondering
- Is the working correct?
- How to interpret the final result. This expectation does not include the variance in $C(t)$ but does this matter? What does this mean if we take a simple form (i.e. $f(C(t),theta) = theta C(t) - C(t)^2$)?
ordinary-differential-equations probability-theory expected-value sde
asked Jan 7 at 22:25
TomTom
436
$endgroup$
add a comment |
$begingroup$
I am currently working through a problem and am having trouble interpreting my results.
I start with a system whose dynamics are described by the equation:
$$
frac{dC}{dt} = f(C(t),theta)
$$
where $theta$ is a random variable independent of time meaning $C(t)$ too becomes a random variable. We can write the equation for the change in the first moment $E[C(t)]$ as:
$$
frac{dE[C(t)]}{dt} = frac{d}{dt}int p_C(C(t)) C(t) dC
$$
where $p_C(C)$ is the probability distribution function for C. Recalling that
$$
E[g(x)] = int g(x) p_x(x) dx
$$
and rearanging the first equation
$$
g(theta) = C(t) = int f(C(t),theta) dt
$$
we can combine to get
begin{align}
frac{dE[C(t)]}{dt} &= frac{d}{dt}int p_{theta}(theta) g(theta) dtheta \
&= frac{d}{dt}int p_{theta}(theta) int f(C(t),theta) dt dtheta \
&= int p_{theta}(theta)f(C(t),theta) dtheta \
&= E[f(C(t),theta)]
end{align}
I am wondering
- Is the working correct?
- How to interpret the final result. This expectation does not include the variance in $C(t)$ but does this matter? What does this mean if we take a simple form (i.e. $f(C(t),theta) = theta C(t) - C(t)^2$)?
ordinary-differential-equations probability-theory expected-value sde
asked Jan 7 at 22:25
TomTom
436
$endgroup$
I am currently working through a problem and am having trouble interpreting my results.
I start with a system whose dynamics are described by the equation:
$$
frac{dC}{dt} = f(C(t),theta)
$$
where $theta$ is a random variable independent of time meaning $C(t)$ too becomes a random variable. We can write the equation for the change in the first moment $E[C(t)]$ as:
$$
frac{dE[C(t)]}{dt} = frac{d}{dt}int p_C(C(t)) C(t) dC
$$
where $p_C(C)$ is the probability distribution function for C. Recalling that
$$
E[g(x)] = int g(x) p_x(x) dx
$$
and rearanging the first equation
$$
g(theta) = C(t) = int f(C(t),theta) dt
$$
we can combine to get
begin{align}
frac{dE[C(t)]}{dt} &= frac{d}{dt}int p_{theta}(theta) g(theta) dtheta \
&= frac{d}{dt}int p_{theta}(theta) int f(C(t),theta) dt dtheta \
&= int p_{theta}(theta)f(C(t),theta) dtheta \
&= E[f(C(t),theta)]
end{align}
I am wondering
- Is the working correct?
- How to interpret the final result. This expectation does not include the variance in $C(t)$ but does this matter? What does this mean if we take a simple form (i.e. $f(C(t),theta) = theta C(t) - C(t)^2$)?
ordinary-differential-equations probability-theory expected-value sde
ordinary-differential-equations probability-theory expected-value sde
asked Jan 7 at 22:25
TomTom
436
asked Jan 7 at 22:25
TomTom
436
edited Jan 7 at 23:42
Tom
asked Jan 7 at 22:25
TomTom
436
asked Jan 7 at 22:25
TomTom
436
asked Jan 7 at 22:25
TomTom
436
436
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