数学代写|Math56a Stochastic processes

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Math56a课程简介

A stochastic process is a random process which evolves with time. The basic model is the Markov chain. This is a set of “states” together with transition probabilities from one state to another. For example, in simple epidemic models there are only two states: $S=$ “susceptible” and $I=$ “infected.” The probability of going from $S$ to $I$ increases with the size of $I$. In the simplest model The $S \rightarrow I$ probability is proportional to $I$, the $I \rightarrow S$ probability is constant and time is discrete (for example, events happen only once per day). In the corresponding deterministic model we would have a “difference equation.”

Prerequisites 

In a continuous time Markov chain, transition events can occur at any time with a certain probability density. The corresponding deterministic model is a first order differential equation. This includes the “general stochastic epidemic.”

The number of states in a Markov chain is either finite or countably infinite. When the collection of states becomes a continuum, e.g., the price of a stock option, we no longer have a “Markov chain.” We have a more general stochastic process. Under very general conditions we obtain a Wiener process, also known as Brownian motion. The mathematics of hedging implies that stock options should be priced as if they are exactly given by this process. Ito’s formula explains how to calculate (or try to calculate) stochastic integrals which give the long term expected values for a Wiener process.

Math56a Stochastic processes HELP（EXAM HELP， ONLINE TUTOR）

A man is playing two slot machines (call them $A$ and $B$ ). Machine $A$ gives a payoff with a probability of $1 / 6$ and machine $B$ gives a payoff with probability $1 / 16$. The man starts by picking a machine at random. Then he plays the machine until he has lost twice (not in a row, just in total). Then he switches machines and continues. For example, suppose that his winning (1) and losing (0) sequence might be:
$$
{ }0 1_1 0_2 0_3 1_4 0_5 1_6 0_7 1_8 0_9 1{10} 0_{11} 0_{12}
$$
Then he will switch machines after $n=2$ since he lost twice. (He switches in the time between $n=2$ and $n=3$ ). He switches back after $n=6$ and then again after $n=10$.

To find the invariant distribution, we need to set up a system of equations based on the probabilities of being in each state. Let $P_n$ denote the probability of being in state $n$, where state $n$ represents playing on machine $A$ ($n$ odd) or machine $B$ ($n$ even) after $n-1$ plays.

We first note that $P_1 = P_2 = 1/2$, since the man starts by picking a machine at random.

For $n \geq 3$, we have:

$$P_n = \frac{1}{2}P_{n-1}p_{n-1} + \frac{1}{2}P_{n-2}(1-p_{n-2})$$

The first term on the right-hand side represents the probability of being in state $n-1$ after $n-2$ plays, and then transitioning to state $n$ by playing on machine $A$ with probability $p_{n-1}$. The second term represents the probability of being in state $n-2$ after $n-3$ plays, and then transitioning to state $n$ by switching from machine $B$ to machine $A$ with probability $1-p_{n-2}$.

Note that $p_{n-1}$ and $1-p_{n-2}$ are given by:

$$p_{n-1} = \begin{cases} \frac{1}{6} & n \text{ odd} \ 0 & n \text{ even} \end{cases}$$

$$1-p_{n-2} = \begin{cases} 0 & n \text{ odd} \ \frac{15}{16} & n \text{ even} \end{cases}$$

Substituting these expressions into the equation for $P_n$ and simplifying, we obtain:

$$P_n = \begin{cases} \frac{1}{2}P_{n-1}\frac{1}{6} + \frac{1}{2}P_{n-2} & n \text{ odd} \ \frac{1}{2}P_{n-1} + \frac{1}{2}P_{n-2}\frac{15}{16} & n \text{ even} \end{cases}$$

Simplifying further, we obtain:

$$P_n = \begin{cases} \frac{1}{12}P_{n-1} + \frac{1}{2}P_{n-2} & n \text{ odd} \ \frac{1}{2}P_{n-1} + \frac{15}{32}P_{n-2} & n \text{ even} \end{cases}$$

We can write these equations in matrix form as:

$$\begin{pmatrix} P_{2m+1} \ P_{2m+2} \end{pmatrix} = \begin{pmatrix} \frac{1}{2} & \frac{1}{12} \ \frac{1}{2} & \frac{15}{32} \end{pmatrix} \begin{pmatrix} P_{2m-1} \ P_{2m} \end{pmatrix}$$

where $m \geq 1$.

Let $X_n$ be a random variable representing the machine the man is playing on at time $n$. We want to find the long-run probability that $X_n$ is equal to machine $B$, which has a higher probability of giving a payoff.

Let $q = \frac{1}{16}$ be the probability of a payoff on machine $B$, and let $p = \frac{1}{6}$ be the probability of a payoff on machine $A$. Let $P_B$ be the long-run probability of playing on machine $B$, so $P_A = 1 – P_B$ is the long-run probability of playing on machine $A$.

Using the law of total probability, we can write:

$$P_B = \sum_{k=1}^\infty P(X_{2k} = B)P(X_1 = A) + \sum_{k=1}^\infty P(X_{2k+1} = B)P(X_1 = B)$$

The first sum represents the probability of playing on machine $B$ at the beginning of each even numbered period, and the second sum represents the probability of playing on machine $B$ at the beginning of each odd numbered period.

Using the fact that the man switches machines after losing twice, we can calculate these probabilities as follows:

$$P(X_{2k} = B)P(X_1 = A) = \left(1 – p^2\right)^{k-1}p$$

The probability that the man switches from machine $A$ to machine $B$ at the beginning of period $2k$ is $(1-p^2)^{k-1}p$, since the man needs to win at least one payoff on machine $A$ in each of the previous $k-1$ periods, and then lose twice in a row in period $2k-1$. Once the man switches to machine $B$, the probability of playing on machine $B$ for the rest of the period is $q$.

Similarly, we have:

$$P(X_{2k+1} = B)P(X_1 = B) = \left(1 – q^2\right)^{k-1}q$$

The probability that the man switches from machine $B$ to machine $A$ at the beginning of period $2k+1$ is $(1-q^2)^{k-1}q$, since the man needs to win at least one payoff on machine $B$ in each of the previous $k$ periods, and then lose twice in a row in period $2k$. Once the man switches to machine $A$, the probability of playing on machine $B$ for the rest of the period is $p$.

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