Intuition Behind Restricted Boltzmann Machine (RBM)

Ho seguito il corso sulle reti neurali di Geoff Hinton su Coursera e anche attraverso l' introduzione alle macchine boltzmann riservate , ma non capivo l'intuizione dietro gli RBM.

Perché dobbiamo calcolare l'energia in questa macchina? E a che serve la probabilità in questa macchina? Ho visto anche questo video . Nel video, ha appena scritto le equazioni di probabilità ed energia prima delle fasi di calcolo e non sembra usarlo da nessuna parte.

In aggiunta a quanto sopra, non sono sicuro a cosa serva la funzione di probabilità?

Gli RBM sono una bestia interessante. Per rispondere alla tua domanda e fare jogging sulla mia memoria, trarrò gli RBM e parlerò attraverso la derivazione. Hai detto che sei confuso sulla probabilità, quindi la mia derivazione sarà dal punto di vista del tentativo di massimizzare la probabilità. Quindi cominciamo.

Gli RBM contengono due diversi insiemi di neuroni, visibili e nascosti, li indicherò rispettivamente e h . Data una configurazione specifica di v e h , lo mappiamo lo spazio di probabilità.$v$$h$$v$$h$

$$p(v,h) = \frac{e^{-E(v,h)}}{Z}$$

Ci sono un paio di cose in più da definire. La funzione surrogata che utilizziamo per mappare da una configurazione specifica allo spazio di probabilità è chiamata funzione energetica . La costante Z è un fattore di normalizzazione per garantire che effettivamente mappiamo allo spazio di probabilità. Ora andiamo a quello che stiamo davvero cercando; la probabilità di un insieme di neuroni visibili, in altre parole, la probabilità dei nostri dati. Z = ∑ v ∈ V ∑ h ∈ H e - E ( v , h ) p ( v $E(v,h)$$Z$

$$Z = \sum_{v \in V}\sum_{h \in H}e^{-E(v,h)}$$ $$p(v)=\sum_{h \in H}p(v,h)=\frac{\sum_{h \in H}e^{-E(v,h)}}{\sum_{v \in V}\sum_{h \in H}e^{-E(v,h)}}$$

Sebbene ci siano molti termini in questa equazione, si riduce semplicemente a scrivere le equazioni di probabilità corrette. Speriamo che, finora, questo ha aiutato a capire il motivo per cui abbiamo bisogno funzione energia per calcolare la probabilità, o ciò che viene fatto più spesso il non normalizzato probabilità . La probabilità non normalizzata viene utilizzata perché la funzione di partizione $p(v)*Z$$Z$ è molto costosa da calcolare.

Ora passiamo alla fase di apprendimento effettiva degli RBM. Per massimizzare la probabilità, per ogni punto dati, dobbiamo fare un gradiente per rendere . Per ottenere le espressioni gradiente ci vogliono alcune acrobazie matematiche. La prima cosa che facciamo è prendere il registro di p ( v ) . D'ora in poi opereremo nello spazio delle probabilità dei log per rendere fattibile la matematica.$p(v)=1$$p(v)$

Prendiamo il gradiente rispetto a i paremeters in p ( v

$$\log(p(v))=\log[\sum_{h \in H}e^{-E(v,h)}]-\log[\sum_{v \in V}\sum_{h \in H}e^{-E(v,h)}]$$ $p(v)$

$$\begin{align} \frac{\partial \log(p(v))}{\partial \theta}=& -\frac{1}{\sum_{h' \in H}e^{-E(v,h')}}\sum_{h' \in H}e^{-E(v,h')}\frac{\partial E(v,h')}{\partial \theta}\\ & + \frac{1}{\sum_{v' \in V}\sum_{h' \in H}e^{-E(v',h')}}\sum_{v' \in V}\sum_{h' \in H}e^{-E(v',h')}\frac{\partial E(v,h)}{\partial \theta} \end{align}$$

Ora l'ho fatto su carta e ho scritto l'equazione semifinale per non perdere molto spazio su questo sito. Ti consiglio di derivare queste equazioni da solo. Ora scriverò alcune equazioni che aiuteranno a continuare la nostra derivazione. Nota che: , p ( v ) = ∑ h ∈ H p ( v , h ) e che p ( h | v ) $Zp(v,h)=e^{-E(v,h')}$$p(v)=\sum_{h \in H}p(v,h)$$p(h|v) = \frac{p(v,h)}{p(h)}$

$$\begin{align} \frac{\partial log(p(v))}{\partial \theta}&= -\frac{1}{p(v)}\sum_{h' \in H}p(v,h')\frac{\partial E(v,h')}{\partial \theta}+\sum_{v' \in V}\sum_{h' \in H}p(v',h')\frac{\partial E(v',h')}{\partial \theta}\\ \frac{\partial log(p(v))}{\partial \theta}&= -\sum_{h' \in H}p(h'|v)\frac{\partial E(v,h')}{\partial \theta}+\sum_{v' \in V}\sum_{h' \in H}p(v',h')\frac{\partial E(v',h')}{\partial \theta} \end{align}$$

And there we go, we derived maximum likelihood estimation for RBM's, if you want you can write the last two terms via expectation of their respective terms (conditional, and joint probability).

Notes on energy function and stochasticity of neurons.

As you can see above in my derivation, I left the definition of the energy function rather vague. And the reason for doing that is that many different versions of RBM implement various energy functions. The one that Hinton describes in the lecture linked above, and shown by @Laurens-Meeus is:

$$E(v,h)=−a^Tv−b^Th−v^TWh.$$

It might be easier to reason about the gradient terms above via the expectation form.

$$\frac{\partial \log(p(v))}{\partial \theta}= -\mathop{\mathbb{E}}_{p(h'|v)}\frac{\partial E(v,h')}{\partial \theta}+\mathop{\mathbb{E}}_{p(v',h')}\frac{\partial E(v',h')}{\partial \theta}$$

The expectation of the first term is actually really easy to calculate, and that was the genius behind RBMs. By restricting the connection the conditional expectation simply becomes a forward propagation of the RBM with the visible units clamped. This is the so called wake phase in Boltzmann machines. Now calculating the second term is much harder and usually Monte Carlo methods are utilized to do so. Writing the gradient via average of Monte Carlo runs:

$$\frac{\partial \log(p(v))}{\partial \theta}\approx -\langle \frac{\partial E(v,h')}{\partial \theta}\rangle_{p(h'|v)}+\langle\frac{\partial E(v',h')}{\partial \theta}\rangle_{p(v',h')}$$

Calculating the first term is not hard, as stated above, therefore Monte-Carlo is done over the second term. Monte Carlo methods use random successive sampling of the distribution, to calculate the expectation (sum or integral). Now this random sampling in classical RBM's is defined as setting a unit to be either 0 or 1 based on its probability stochasticly, in other words, get a random uniform number, if it is less than the neurons probability set it to 1, if it is greater than set it to 0.

In addition to the existing answers, I would like to talk about this energy function, and the intuition behind that a bit. Sorry if this is a bit long and physical.

The energy function describes a so-called Ising model, which is a model of ferromagnetism in terms of statistical mechanics / quantum mechanics. In statistical mechanics, we use a so-called Hamiltonian operator to describe the energy of a quantum-mechanical system. And a system always tries to be in the state with the lowest energy.

Now, the Ising model basically describes the interaction between electrons with a spin $\sigma_k$ of either +1 or -1, in presence of an external magnetic field $h$. The interaction between two electrons $i$ and $j$ is described by a coefficient $J_{ij}$. This Hamiltonian (or energy function) is

$$\hat{H} = \sum_{i,j} J_{ij} \sigma_i \sigma_j - \mu \sum_j h_j \sigma_j$$ where $\hat{H}$ denotes the Hamiltonian. A standard procedure to get from an energy function to the probability, that a system is in a given state (i.e. here: a configuration of spins, e.g. $\sigma_1 = {+1}, \sigma_2 = {-1}, ...$) is to use the Boltzmann distribution, which says that at a temperature $T$, the probability $p_i$ of the system to be in a state $i$ with energy $E_i$ is given by $$p_i = \frac{\exp(-E_i/kT)}{\sum_{i}\exp(-E_i/kt)}$$ At this point, you should recognize that these two equations are the exact same equations as in the videos by Hinton and the answer by Armen Aghajanyan. This leads us to the question:

What does the RBM have to do with this quantum-mechanical model of ferromagnetism?

We need to use a final physical quantity: the entropy. As we know from thermodynamics, a system will settle in the state with the minimal energy, which also corresponds to the state with the maximal entropy.

As introduced by Shanon in 1946, in information theory, the entropy $H$ can also be seen as a measure of the information content in $X$, given by the following sum over all possible states of $X$:

$$H(X) = -\sum_i P(x_i) \log P(x_i)$$ Now, the most efficient way to encode the information content in $X$, is to use a way that maximizes the entropy $H$.

Finally, this is where we get back to RBMs: Basically, we want this RBM to encode as much information as possible. So, as we have to maximize the (information-theoretical) entropy in our RBM-System. As proposed by Hopfield in 1982, we can maximize the information-theoretical entropy exactly like the physical entropy: by modelling the RBM like the Ising model above, and use the same methods to minimize the energy. And that is why we need this strange energy function for in an RBM!

The nice mathematical derivation in Armen Aghajanyan's answer shows everything we need to do, to minimize the energy, thus maximizing entropy and storing / saving as much information as possible in our RBM.

_{PS: Please, dear physicists, forgive any inaccuracies in this engineer's derivation. Feel free to comment on or fix inaccuracies (or even mistakes).}

The answer of @Armen has gave myself a lot of insights. One question hasn't been answered however.

The goal is to maximize the probability (or likelihood) of the $v$. This is correlated to minimizing the energy function related to $v$ and $h$:

$$E(v,h) = -a^{\mathrm{T}} v - b^{\mathrm{T}} h -v^{\mathrm{T}} W h$$

Our variables are $a$, $b$ and $W$, which have to be trained. I'm quite sure this training will be the ultimate goal of the RBM.