Sisi Tang – Page 4 – A Quantitative Finance Website

March 8, 2019March 8, 2021

European call option v.s. Asian call option

Let a stock follow a Geometric Brownian motion

$\textnormal{d}S_t = S_t(r \textnormal{d} t + \sigma \textnormal{d} W_t)$

with constant $r,\sigma > 0$ . Let

$A_T = \frac{1}{T}\int_0^T S_u \textnormal{d} u$

European Asian call option has the payoff $(A_T-K)^+$ and European vanilla call option has the payoff $(S_T-K)^+$ . Then European vanilla option has higher value.

Proof.

By (1)Jensen’s inequality, (2)Fubini, and (3)the fact that the longer the maturity is the higher the vanilla European option’s value is, we have

$\begin{aligned} &~ \mathbb{E}[e^{-rT}(A_T-K)^+] \\= &~ \mathbb{E}[e^{-rT}(\frac{1}{T}\int_0^T S_u \textnormal{d} u -K)^+] \\\leq & ~ \mathbb{E}[e^{-rT}\frac{1}{T}\int_0^T (S_u-K)^+ \textnormal{d} u] \\= &~ \frac{1}{T}\int_0^T \mathbb{E}[e^{-rT} (S_u-K)^+ ] \textnormal{d} u \\\leq &~ \frac{1}{T}\int_0^T \mathbb{E}[e^{-ru} (S_u-K)^+ ] \textnormal{d} u \\\leq &~ \frac{1}{T}\int_0^T \mathbb{E}[e^{-rT} (S_T-K)^+ ] \textnormal{d} u \\ = &~ \mathbb{E}[e^{-rT} (S_T-K)^+ ] \\ \end{aligned}$

Notice that we didn’t use the dynamics of Geometric Brownian motion. Above deduction is true as long as (3) is true.

February 12, 2019August 3, 2019

Neural Network

This post is my study notes of Andrew Ng’s course. https://www.andrewng.org/courses/

Neural network is a convolution of several logistic regressions. It allows some dependence between those regressions. Neural network incorporates more coefficients that will be learned from the date, so it should provide higher accuracy than a single logistic regression. The only thing we need to pay attention is over-fitting.

Here we use neural network with 3 layers (an input layer, a hidden layer, and an output layer) as an example for background information. The case of more layers is quite similar. In this article, our inputs are 25 by 25 pixels images. Since the images are of size $25 \times 25$ , this gives us $625$ input layer units (not counting the extra bias unit). The training data will be loaded into the variables $X$ and $y$ , where $X$ is the image, and $y$ is the label.

Let $m$ be the number of inputs(images in our case), and $K$ be the number of possible lables. The cost function for the neural network (without regularization) is

$\begin{aligned}J(\theta) = \frac{1}{m} \sum_{i=1}^{m} \sum_{k=1}^K &[-y_k^{(i)}\log(h_{\theta}(x^{(i)})k) \\ &- (1-y_k^{(i)}) \log(1-(h{\theta}(x^{(i)})_k)) ]\end{aligned}$

To avoid over-fitting, we use the cost function for neural networks with regularization

$\begin{aligned}J(\theta) = \frac{1}{m} \sum_{i=1}^{m} \sum_{k=1}^K [-y_k^{(i)}\log(h_{\theta}(x^{(i)})k) \\ - (1-y_k^{(i)}) \log(1-(h{\theta}(x^{(i)})k)) ] \\+\frac{\lambda}{2m} [ \sum_{j=1}^{25} \sum_{k=1}^{625} (\Theta_{j,k}^{(1)})^2 + \sum_{j=1}^{2}\sum_{k=1}^{25} (\Theta_{j,k}^{(2)})^2 ]\end{aligned}$

When training neural networks, it is important to randomly initialize the parameters for symmetry breaking. One effective strategy for random initialization is to randomly select values for $\Theta^{(l)}$ uniformly in the range $[-\epsilon,\epsilon]$ . We use $\epsilon=0.12$ . This range of values ensures that the parameters are kept small and makes the learning more efficient.

One effective strategy for choosing $\epsilon$ is to base it on the number of units in the network. A good choice of $\epsilon$ is $\epsilon = \frac{\sqrt{6}}{\sqrt{L_{in}+L_{out}}}$ , where $L_{in} = sl$ and $L_{out} = sl+1$ are the number of units in the layers adjacent to $\Theta^{(l)}$ .

The error of the neural network is obtained by the backpropagation algorithm. The intuition behind the backpropagation algorithm is as follows. Given a training example $(x(t), y(t))$ , we will first run a “forward pass” to compute all the activations throughout the network, including the output value of the hypothesis $h_{\Theta}(x)$ . Then, for each node $j$ in layer $l$ , we would like to compute
an error term $\delta_j^{(l)}$ that measures how much that node was responsible for any errors in our output.

For an output node, we can directly measure the difference between the network’s activation and the true target value, and use that to define $\delta^{(3)}_j$ (since layer 3 is the output layer). For the hidden units, we can compute $\delta^{(l)}_j$ based on a weighted average of the error terms of the nodes in layer $(l + 1)$ .

Procedure

In detail, here is the backpropagation algorithm. We should implement steps 1 to 4 in a loop that processes one example at a time. Concretely, we should implement a for-loop for $t=1:m$ and place steps 1-4 below inside the for-loop, with the t-th iteration performing the calculation on the t-th training example $(x(t), y(t))$ . Step 5 will divide the accumulated gradients by $m$ to obtain the gradients for the neural network cost function.

Set the input layer’s values $(a(1))$ to the t-th training example $x(t)$ . Perform a feedforward pass (Figure ??), computing the activations $(z(2); a(2); z(3); a(3))$ for layers 2 and 3. Note that we need to add $a+1$ term to ensure that the vectors of activations for layers $a(1)$ and $a(2)$ also include the bias unit.
For each output unit $k$ in layer 3 (the output layer), set
$\delta_k^{(3)} = (a^{(3)}_k - y_k)$
where $y_k \in [0,1]$ indicates whether the current training example belongs to class k ( $y_k = 1, k = 1,2$ ), or if it belongs to a different class ( $y_k = 0$ ).
For hidden layer $l=2$ , set
$\delta^{(2)} = (\Theta^{(2)})^T \delta^{(3)} .* g'(z^{(2)}).$
Accumulate the gradient from this example using the following formula. Note that we should skip or remove $\delta^{(2)}_0$ .
$\Delta^{(l)} = \Delta^{(l)} + \delta^{(l)}(a^{(l)})^T$
Obtain the (unregularized) gradient for the neural network cost function by dividing the accumulated gradients by $m$ :
$\frac{\partial}{\partial \Theta_{ij}^{(l)}} J(\Theta) = D_{ij}^{(l)} = \frac{1}{m} \Delta_{ij}^{(l)}$

To account for regularization, it turns out that we can add this as an additional term after computing the gradients using backpropagation. Specifically, after we have computed $\Delta^{(l)}_{ij}$ using backpropagation, we should add regularization using

$\frac{\partial}{\partial \Theta_{ij}^{(l)}} J(\Theta) = D_{ij}^{(l)} = \frac{1}{m} \Delta_{ij}^{(l)},\quad \mbox{for} j = 0.$

$\frac{\partial}{\partial \Theta_{ij}^{(l)}} J(\Theta) = D_{ij}^{(l)} = \frac{1}{m} \Delta_{ij}^{(l)} + \frac{\lambda}{m}\Theta_{ij}^{(l)}, \quad \mbox{for} j \geq 1.$

After we have successfully implemented the neural network cost function and gradient computation by feedforward propagation and backpropagation, the next step will be learning a good set of parameters by minimizing the cost function. Since the cost function is not convex, there is no guarantee that we can always find the global minimum. But we should try to increase the number of iterations in our minimizer(say gradient descent, or conjugate descent). And perform the solver several times since the initialization is random, and different initialization may results in different local minimum.

February 4, 2019March 8, 2021

Cross Hedging

When our study group read John Hull’s Options, Futures, and Other Derivatives 10th Edition book section 3.4 Cross Hedging, the hedging ratio $h^*$ was given directly in (3.1). We filled in the derivation of it here.

Goal

We want to hedge an asset using a future contract whose underlying asset is different from the one being hedged. This usually happens for example in the case that the asset being hedged is not available in the future market. If we are going to sell an asset, then we take long position in a future contract; short otherwise.

Setting

$h$ : hedge ratio. The ratio of the size of the position taken in futures contracts to the size of the exposure.
$\Delta S$ : change of spot price of the asset to be hedged.
$\Delta F$ : change of price in the future contract.

Formulation

Mathematically, we want to minimize the variance of our portfolio value, i.e.,

$\begin{aligned}\mbox{Var}(\Delta S-h\Delta F) &= \mathbf{E}[(\Delta S-\overline{\Delta S}-h(\Delta F-\overline{\Delta F}))^2] \\&= \mathbf{E}[(\Delta S-\overline{\Delta S})^2] + h^2\mathbf{E}[(\Delta F-\overline{\Delta F})^2] \\ & \quad - 2h\mathbf{E}[(\Delta S-\overline{\Delta S})(\Delta F-\overline{\Delta F})] \\&= \sigma_{\Delta S}^2+ h^2\sigma_{\Delta F}^2 - 2h~\mbox{cov}(\Delta S,\Delta F) ,\end{aligned}$

where $\sigma_{\Delta S}, \sigma_{\Delta F}$ denote the standard deviation of $\Delta S$ , standard deviation of $\Delta F$ respectively. This is a quadratic fucntion in $h$ . The minimum of it is achieved at

$h=\frac{\mbox{cov}(\Delta S,\Delta F)}{\sigma_{\Delta F}^2}=\rho\frac{\sigma_{\Delta S}}{\sigma_{\Delta F}},$

where $\rho$ is the correlation of $\Delta S$ and $\Delta F$ . This formula for $h$ is exactly the slope of linear regression $\Delta S \sim \Delta F$ . Our derivation here also explains why linear regression is the minimum variance estimator.

Then, the variance

$\mbox{Var}(\Delta S-h\Delta F)=(1-\rho^2)\sigma_{\Delta S}^2.$