- #1

perplexabot

Gold Member

- 329

- 5

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

[tex] \phi_{k+1}=\phi_{k}+v_k [/tex] where, [itex]v_k\overset{iid}{\sim}\mathcal{N}(0,\sigma^2)[/itex] and [itex]\phi_{k+1}[/itex] be a scalar.

Let us find the following first two conditional moments

[tex]

\begin{equation*}

\begin{split}

E[\phi_{k+1}|\phi_k] &= \phi_k \\

cov[\phi_{k+1}|\phi_k] &= E[(\phi_{k+1}-\phi_k)(\phi_{k+1}-\phi_k)^T] = E[v_k^2] = \sigma^2

\end{split}

\end{equation*}

[/tex] Where we know [itex]p(\phi_{k+1}|\phi_k)[/itex] is a normal distribution, finding [itex]log[p(\phi_{k+1}|\phi_k)][/itex]

[tex]

log[p(\phi_{k+1}|\phi_k)] = \frac{-1}{2\sigma^2}(\phi_{k+1}-\phi_{k})^2

[/tex]

I need to find the derivative (with respect to [itex]\phi_k[/itex]) of [itex]log[p(\phi_{k+1}|\phi_k)][/itex].

Finally, my question... When I find the derivative of this quantity, do I need to substitute for [itex]\phi_{k+1}[/itex] such that

[tex]

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

[/tex]

This will end up giving me zero if I take the derivative with respect to [itex]\phi_k[/itex] (right? or is this just telling me that I can substitute [itex]v_k[/itex] for [itex]\phi_{k+1}-\phi_k[/itex]... 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 [itex]\phi_{k+1}[/itex] is a function of [itex]\phi_k[/itex]

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.