perplexabot
Gold Member
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Hey all,
I've had this point of confusion for a bit and I have thought that with time I may be able to clear it out myself. Nope, hasn't happened. I think I need help.
Let us say we have the following
\phi_{k+1}=\phi_{k}+v_k where, v_k\overset{iid}{\sim}\mathcal{N}(0,\sigma^2) and \phi_{k+1} be a scalar.
Let us find the following first two conditional moments
<br /> \begin{equation*}<br /> \begin{split}<br /> E[\phi_{k+1}|\phi_k] &= \phi_k \\<br /> cov[\phi_{k+1}|\phi_k] &= E[(\phi_{k+1}-\phi_k)(\phi_{k+1}-\phi_k)^T] = E[v_k^2] = \sigma^2 <br /> \end{split}<br /> \end{equation*}<br /> Where we know p(\phi_{k+1}|\phi_k) is a normal distribution, finding log[p(\phi_{k+1}|\phi_k)]
<br /> log[p(\phi_{k+1}|\phi_k)] = \frac{-1}{2\sigma^2}(\phi_{k+1}-\phi_{k})^2<br />
I need to find the derivative (with respect to \phi_k) of log[p(\phi_{k+1}|\phi_k)].
Finally, my question... When I find the derivative of this quantity, do I need to substitute for \phi_{k+1} such that
<br /> log[p(\phi_{k+1}|\phi_k)] = \frac{-1}{2\sigma^2}((\phi_k+v_k)-\phi_k)^2 = \frac{-1}{2\sigma^2}(v_k)^2<br />
This will end up giving me zero if I take the derivative with respect to \phi_k (right? or is this just telling me that I can substitute v_k for \phi_{k+1}-\phi_k... i have no clue... if i do that, then i am back to where i started and i will forever be stuck in a loop of substition, LOL).
On the other hand, I could NOT do that substitution and up with a non zero answer. But that is kind of weird if I do that (not substituting), I am basically disregarding the fact that \phi_{k+1} is a function of \phi_k
I find this really confusing. What is the correct way to do this? Please help me clear this problem out as it has been an issue for a while : /
Thank you for reading.
I've had this point of confusion for a bit and I have thought that with time I may be able to clear it out myself. Nope, hasn't happened. I think I need help.
Let us say we have the following
\phi_{k+1}=\phi_{k}+v_k where, v_k\overset{iid}{\sim}\mathcal{N}(0,\sigma^2) and \phi_{k+1} be a scalar.
Let us find the following first two conditional moments
<br /> \begin{equation*}<br /> \begin{split}<br /> E[\phi_{k+1}|\phi_k] &= \phi_k \\<br /> cov[\phi_{k+1}|\phi_k] &= E[(\phi_{k+1}-\phi_k)(\phi_{k+1}-\phi_k)^T] = E[v_k^2] = \sigma^2 <br /> \end{split}<br /> \end{equation*}<br /> Where we know p(\phi_{k+1}|\phi_k) is a normal distribution, finding log[p(\phi_{k+1}|\phi_k)]
<br /> log[p(\phi_{k+1}|\phi_k)] = \frac{-1}{2\sigma^2}(\phi_{k+1}-\phi_{k})^2<br />
I need to find the derivative (with respect to \phi_k) of log[p(\phi_{k+1}|\phi_k)].
Finally, my question... When I find the derivative of this quantity, do I need to substitute for \phi_{k+1} such that
<br /> log[p(\phi_{k+1}|\phi_k)] = \frac{-1}{2\sigma^2}((\phi_k+v_k)-\phi_k)^2 = \frac{-1}{2\sigma^2}(v_k)^2<br />
This will end up giving me zero if I take the derivative with respect to \phi_k (right? or is this just telling me that I can substitute v_k for \phi_{k+1}-\phi_k... i have no clue... if i do that, then i am back to where i started and i will forever be stuck in a loop of substition, LOL).
On the other hand, I could NOT do that substitution and up with a non zero answer. But that is kind of weird if I do that (not substituting), I am basically disregarding the fact that \phi_{k+1} is a function of \phi_k
I find this really confusing. What is the correct way to do this? Please help me clear this problem out as it has been an issue for a while : /
Thank you for reading.