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Magazine Chapter 10: Problem 18
Show that \(\\{Y(t), t \geqslant 0\\}\) is a Martingale when $$ Y(t)=B^{2}(t)-t $$ What is \(E[Y(t)] ?\) Hint: First compute \(E[Y(t) \mid B(u), 0 \leqslant u \leqslant s]\).
Short Answer
Expert verified
In summary, the process \(Y(t) = B^2(t) - t\) is a martingale, as it satisfies the necessary conditions. The expected value \(E[Y(t)]\) is 0.
Step by step solution
01
Check Condition 1 - Finite expected value for all t
By definition, \(B(t)\) is a standard Brownian motion. Therefore, \(B(t)\) has expected value \(E[B(t)] = 0\) and its variance is \(Var(B(t)) = t\). So, \(E[B^2(t)] = t\), because it is a well-known property of Brownian motion. Now, we compute the expected value of \(Y(t)\):
\[ E[Y(t)] = E[B^2(t) - t] \]
Since the expected value operator is linear, we can rewrite the equation as:
\[ E[Y(t)] = E[B^2(t)] - E[t] \]
Given that we already know \(E[B^2(t)] = t\), we can easily find:
\[ E[Y(t)] = t - t = 0 \]
Thus, \(E[Y(t)]\) is indeed finite, and condition 1 holds.
02
Check Condition 2 - Conditional expectation
For all \(0 \leq s \leq t\), we need to verify if the following condition holds:
\[ E[Y(t) \mid B(u), 0 \leq u \leq s] = Y(s) \]
First, it's important to note that since \(B(t)\) is a martingale, then \(B(t) - B(s)\) is independent of \(B(u)\) for \(0 \leq u \leq s\). Now, let's calculate the conditional expectation:
\[ E[Y(t) \mid B(u), 0 \leq u \leq s] = E[(B^2(t) - t) \mid B(u), 0 \leq u \leq s] \]
Now we'll exploit the independence between the increments of Brownian motion, and apply the law of total expectation:
\[ E[Y(t) \mid B(u), 0 \leq u \leq s] = E[((B(t) - B(s)) + B(s))^2 - t) \mid B(u), 0 \leq u \leq s] \]
\[ E[Y(t) \mid B(u), 0 \leq u \leq s] = E[((B(t) - B(s))^2 + 2B(s)(B(t) - B(s)) + B(s)^2 - t) \mid B(u), 0 \leq u \leq s] \]
Use linearity of conditional expectation:
\[ E[Y(t) \mid B(u), 0 \leq u \leq s] = E[(B(t) - B(s))^2 \mid B(u), 0 \leq u \leq s] + 2B(s)E[B(t) - B(s) \mid B(u), 0 \leq u \leq s] \\ + E[B(s)^2 \mid B(u), 0 \leq u \leq s]- E[t \mid B(u), 0 \leq u \leq s] \]
Since \(B(t) - B(s)\) is independent of \(B(u)\) for \(0 \leq u \leq s\):
\[ E[Y(t) \mid B(u), 0 \leq u \leq s] = E[(B(t) - B(s))^2] + 2B(s)E[B(t) - B(s)] + B(s)^2 - t \]
As mentioned previously, we know that \(E[B(t) - B(s)] = 0\), so now we just need the expected value of the squared increment:
\[ E[(B(t) - B(s))^2] = Var(B(t) - B(s)) = t - s \]
Now we can compute the conditional expectation:
\[ E[Y(t) \mid B(u), 0 \leq u \leq s] = (t - s) + 0 + B(s)^2 - t \]
\[ E[Y(t) \mid B(u), 0 \leq u \leq s] = B(s)^2 - s = Y(s) \]
So, \(Y(t)\) satisfies condition 2. Since both conditions are met, we can conclude that \(Y(t)\) is a martingale.
03
Calculate E[Y(t)]
Now we will compute the expected value of Y(t) which we already found in Step 1:
\[E[Y(t)] = 0\]
So, the expected value of the process \(Y(t)=B^2(t) - t\) is 0.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Brownian motion
Brownian motion is a fundamental concept in probability theory and stochastic processes. It describes the random continuous motion of particles suspended in a fluid, but it is also used to model a variety of random phenomena.
- **Random Walk:** At its core, Brownian motion can be thought of as a random walk with continuous movement, meaning that the path it takes is unpredictable and varies over time.
- **Standard Brownian Motion:** In mathematics, a standard Brownian motion, often denoted as \(B(t)\), has an expected value \(E[B(t)] = 0\). This means that, on average, the position is zero, and its variance \(Var(B(t)) = t\) grows linearly with time.
- **Independent Increments:** The most crucial property is that Brownian motion has independent increments. This means the movement over any time period is independent of any past movements.
Conditional expectation
Conditional expectation is a way to compute the expected value of a random variable given some condition or prior information. It is a crucial concept in probability, playing a key role in determining future expectations.
- **Conceptual View:** Think of conditional expectation, \(E[X|Y]\), as the expected value of a random variable \(X\) given that another random variable \(Y\) is known. It's like asking, "What should we expect given what we already know?"
- **Application in Martingales:** In the context of martingales, conditional expectations are used to verify if a stochastic process remains fair over time. The definition relies heavily on the idea that given past information, the future expectation of the process remains unchanged.
- **Law of Total Expectation:** Often involves breaking down a complex expectation into simpler parts and depends heavily on independence in stochastic processes like Brownian motion.
Expected value
The expected value is a fundamental concept in statistics and probability, representing the long-term average or mean value of a random variable. It provides a single summary statistic about the distribution of possible outcomes.
- **Definition:** For a random variable \(X\), the expected value \(E[X]\) is calculated as the sum (or integral, for continuous variables) of all possible values weighted by their probability.
- **Linear Property:** One essential property of expected value is linearity. If \(X\) and \(Y\) are random variables, then \(E[X+Y] = E[X] + E[Y]\). This property is particularly useful when dealing with combinations of random variables.
- **Role in Martingales:** In martingales, expected value plays a critical role in verifying the fairness of a process over time. A martingale process is defined such that the conditional expected value of future values, given past values, is equal to its current value.
Variance
Variance is a statistical measure that quantifies the dispersion or variability of a set of values. It tells us how much the values differ from the average (expected value).
- **Definition:** For a random variable \(X\), its variance is given by \(Var(X) = E[(X - E[X])^2]\). It measures the average of the squared deviations from their mean.
- **Significance:** Variance helps to understand the spread of the values around the mean. A higher variance means more spread out data, whereas a lower variance indicates data points are close to the mean.
- **Use in Stochastic Processes:** In processes like Brownian motion, variance provides an insight into volatility. For instance, the variance of \(B(t)\), a standard Brownian motion, is \(t\), showing that variance grows linearly with time and indicates the degree of variability of particle movement.
