# Lecutre 9

## GANs

Give up modeling p(x), but allows to draw samples from p(x)

Assume we have data xi drawn from distribution $$p\_{data}(x)$$, we want to sample from $$p\_{data}$$

Idea: Introduce a latent variable z with simple prior p(z). Sample $$z \sim p(z)$$ and pass to Generator Network $$x = G(z)$$. Then x is sample from the generator distribution $$p\_G$$. Want $$p\_G = p\_{data}$$

<figure><img src="/files/hO6nTUW9rxamy2743ldo" alt=""><figcaption><p>GANs</p></figcaption></figure>

Training Objective

<figure><img src="/files/vKdqG0e3SYkdpw11vgNE" alt=""><figcaption><p>Objective</p></figcaption></figure>

Train using alternating gradient updates.

$$= \min\_G \max\_D V(G,D)$$

For t in 1, ..., T:

1. update D. $$D = D + \alpha\_D \frac{\partial V}{\partial D}$$
2. update G. $$G = G - \alpha\_G \frac{\partial V}{\partial G}$$

At start of training, generator is very bad and discriminator can easily tell real/fake so D(G(z)) close to 0.

Problem: vanishing gradient for G.

Solution: G is trained to minimize log(1-D(G(z)). Instead, train G to minimize -log(D(G(z)). Then G gets strong gradient at start of training.

It achieves global minimum when $$p\_G = p\_{data}$$

$$\min\_G \max\_D (E\_{x \sim p\_{data}}\[\log D(x)] + E\_{z \sim p(z)} \[\log (1-D(G(z)))])$$

$$=\min\_G \max\_D (E\_{x\sim p\_{data}}\[\log D(x)] + E\_{x \sim p\_G} \[\log (1-D(x))])$$ Change of variable on second term.

$$= \min\_G \max\_D \int\_x (p\_{data}(x) \log D(x) + p\_G (x) \log (1-D(x))) dx$$ Definition of Expectation.

$$= \min\_G \int\_x max\_D(p\_{data}(x) \log D(x) + p\_G (x) \log (1-D(x))) dx$$ push max\_D inside integral.

$$f(y) = a\log y - b \log (1-y)$$ . a = p\_data (x) y = D(x), b = p\_G(x). (Side computation to compute max)

$$f^\prime (y) = \frac{a}{y} - \frac{b}{1-y}$$, $$f^\prime (y) = 0 \iff y = \frac{a}{a+b}$$ (local max)

Optimal Discriminator: $$D\_G^\* (x) = \frac{p\_{data}(x)}{p\_{data}(x) + p\_G(x)}$$

$$= \min\_G \int\_x (p\_data(x) \log D\_G^\* (x) + p\_G(x) \log (1-D\_G^\*(x)))dx$$

$$= \min\_G \int\_x (p\_data(x) \log \frac{p\_{data}(x)}{p\_{data}(x) + p\_G(x)} + p\_G(x) \log \frac{p\_G(x)}{p\_{data}(x) + p\_G(x)} dx$$

$$= \min\_G (E\_{x \sim p\_{data}} \[\log \frac{p\_{data}(x)}{p\_{data}(x) + p\_G(x)}] + E\_{x\sim p\_G} \[\log \frac{p\_G(x)}{p\_{data}(x) + p\_G(x)}])$$ (definition of expectation)

$$= \min\_G (E\_{x \sim p\_{data}} \[\log \frac{2}{2} \frac{p\_{data}(x)}{p\_{data}(x) + p\_G(x)}] + E\_{x\sim p\_G} \[\log \frac{2}{2} \frac{p\_G(x)}{p\_{data}(x) + p\_G(x)}])$$ (Multiply by a constant)

$$= \min\_G (E\_{x \sim p\_{data}} \[\log \frac{2 \cdot p\_{data}(x)}{p\_{data}(x) + p\_G(x)}] + E\_{x\sim p\_G} \[\log  \frac{2 \cdot p\_G(x)}{p\_{data}(x) + p\_G(x)}] - \log 4)$$

KL Divergence: $$KL(p,q) = E\_{x \sim p} \[\log \frac{p(x)}{q(x)}]$$

$$= \min\_G (KL(p\_{data}, \frac{p\_{data} + p\_G}{2}) + KL(p\_G, \frac{p\_{data} + p\_G}{2}) - \log 4)$$

Jensen-Shannon Divergence: $$JSD(p,q) = \frac{1}{2}KL(p, \frac{p+q}{2}) + \frac{1}{2} KL (q, \frac{p+q}{2})$$

$$=\min\_G (2\cdot JSD(p\_{data}, p\_G) - \log 4)$$

JDS is always nonnegative, and zero only when two distribution are equal, thus p\_data = p\_G is global min.&#x20;

Summary: Global min and max happens when:

1. $$D\_G^\* (x) = \frac{p\_{data}(x)}{p\_{data}(x) + p\_G(x)}$$ (Optimal discriminator for any G)
2. $$p\_G(x) = p\_{data}(x)$$ (Optimal generator for optimal D)

Caveats:

1. G and D are neural nets with fixed architecture, we don't know whether they can actually represent the optimal D and G.
2. This tells nothing about convergence to the optimal solution.

### Conditional GANs

Learn p(x|y) instead of p(y). Make generator and discriminator both take label y as additional input.

<figure><img src="/files/fTm9agVkT7Tt7Cf4vHWc" alt=""><figcaption><p>Batch Normalization</p></figcaption></figure>


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