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Question 1:
Consider a stochastic process {X(t)} defined as a Poisson process with rate λ = 0.5. Find
the probability that exactly 2 events occur in the time interval [0, 4].
Solution: To solve this problem, we need to use the properties of a Poisson process.
The number of events that occur in a fixed interval of time in a Poisson process follows
a Poisson distribution.
The probability mass function (PMF) of a Poisson distribution is given by: P(X = k) =
(e^(-λ) * λ^k) / k!
Where X represents the number of events, λ is the rate, and k is the desired number of
events.
In this case, λ = 0.5 and the time interval is [0, 4]. Since the rate is constant, we can
consider the time interval as [0, t], where t = 4.
The number of events, X, follows a Poisson distribution with rate λt. Therefore, we
have: λt = 0.5 * 4 = 2

To find the probability of exactly 2 events occurring in the time interval [0, 4], we
substitute λ = 2 and k = 2 into the Poisson PMF:
P(X = 2) = (e^(-2) * 2^2) / 2! = (e^(-2) * 4) / 2 = (1/4) * e^(-2)
Calculating this value, we get: P(X = 2) ≈ 0.0902
Therefore, the probability that exactly 2 events occur in the time interval [0, 4] for the
given Poisson process is approximately 0.0902.
Question 2:
A particle is moving along a straight line, and its position at time t is given by X(t) = t^2 +
W(t), where W(t) is a standard Wiener process. Find the mean and variance of X(t).
Solution: To find the mean and variance of X(t), we need to calculate the expected value
(mean) and the second moment (variance) of X(t).
Mean of X(t): The mean of X(t) is given by E[X(t)].

E[X(t)] = E[t^2 + W(t)] = E[t^2] + E[W(t)]
The expected value of t^2 is simply t^2, as t is a deterministic variable.
E[X(t)] = t^2 + E[W(t)]
For a standard Wiener process, the expected value E[W(t)] at any given time t is 0.
E[X(t)] = t^2 + 0 = t^2
Therefore, the mean of X(t) is t^2.
Variance of X(t): The variance of X(t) is given by Var[X(t)].
Var[X(t)] = Var[t^2 + W(t)] = Var[t^2] + Var[W(t)] + 2 * Cov(t^2, W(t))
The variance of t^2 is 0 since t is a deterministic variable.
Var[X(t)] = 0 + Var[W(t)] + 2 * Cov(t^2, W(t))
For a standard Wiener process, the variance Var[W(t)] at any given time t is t.
Var[X(t)] = 0 + t + 2 * Cov(t^2, W(t))
The covariance Cov(t^2, W(t)) can be calculated using the property that Cov(aX, Y) =
a * Cov(X, Y), where a is a constant.

Cov(t^2, W(t)) = t * Cov(t, W(t))
Since t and W(t) are independent, the covariance between them is 0.
Cov(t^2, W(t)) = t * 0 = 0
Therefore, Var[X(t)] = 0 + t + 2 * 0 = t.
The variance of X(t) is t.
In summary: Mean of X(t) = t^2 Variance of X(t) = t
Question 3: Consider a discrete-time Markov chain with the following transition
probability matrix:
P = [[0.8, 0.2], [0.4, 0.6]]
where P[i, j] represents the probability of transitioning from state i to state j. If the
initial state is S1, what is the probability of being in state S2 after 3 time steps?
Solution:
To find the probability of being in state S2 after 3 time steps, we can use the transition
probability matrix and apply the power of matrix multiplication.

Let's denote the initial state vector as V0, where V0 = [1, 0]. This means that the system
starts in state S1 with probability 1 and state S2 with probability 0.
To find the state probabilities after 3 time steps, we multiply the initial state vector V0 by
the transition probability matrix P three times.
V3 = V0 * P^3
Performing the matrix multiplication:
V3 = [1, 0] * P^3 = [1, 0] * P * P * P
Calculating P * P:
P * P = [[0.8, 0.2], [0.4, 0.6]] * [[0.8, 0.2], [0.4, 0.6]] = [[0.8 * 0.8 + 0.2 * 0.4, 0.8 * 0.2 +
0.2 * 0.6], [0.4 * 0.8 + 0.6 * 0.4, 0.4 * 0.2 + 0.6 * 0.6]] = [[0.76, 0.24], [0.52, 0.48]]
Now, we multiply P * P by P:
P^3 = [[0.76, 0.24], [0.52, 0.48]] * [[0.8, 0.2], [0.4, 0.6]] = [[0.76 * 0.8 + 0.24 * 0.4, 0.76
* 0.2 + 0.24 * 0.6], [0.52 * 0.8 + 0.48 * 0.4, 0.52 * 0.2 + 0.48 * 0.6]] = [[0.688, 0.312],
[0.656, 0.344]]
Finally, we calculate V3:
V3 = [1, 0] * [[0.688, 0.312], [0.656, 0.344]] = [1 * 0.688 + 0 * 0.656, 1 * 0.312 + 0 *
0.344] = [0.688, 0.312]
Therefore, the probability of being in state S2 after 3 time steps is approximately 0.312 or
31.2%. For any help regarding Statistics Assignment Help

Question 4: A company receives customer calls at its call center according to a Poisson
process with an average rate of 5 calls per hour. Find the probability that the company
receives at least 8 calls in a 2-hour period.
Solution: To solve this problem, we can use the properties of the Poisson distribution.
The number of calls that the company receives in a 2-hour period follows a Poisson
distribution with a rate of 5 calls per hour. Therefore, the rate parameter for the 2-hour
period is λ = 5 * 2 = 10.
Let X represent the number of calls received in a 2-hour period. We want to find P(X ≥
8).
P(X ≥ 8) = 1 - P(X < 8)
To find P(X < 8), we can sum the probabilities of all possible values from 0 to 7 using
the Poisson probability mass function (PMF):
P(X < 8) = P(X = 0) + P(X = 1) + P(X = 2) + ... + P(X = 7)
The PMF of the Poisson distribution is given by:

P(X = k) = (e^(-λ) * λ^k) / k!
Substituting λ = 10 and summing the probabilities, we have:
P(X < 8) = P(X = 0) + P(X = 1) + P(X = 2) + ... + P(X = 7) = (e^(-10) * 10^0) / 0! +
(e^(-10) * 10^1) / 1! + (e^(-10) * 10^2) / 2! + ... + (e^(-10) * 10^7) / 7!
Calculating this sum, we get:
P(X < 8) ≈ 0.199
Therefore, the probability that the company receives at least 8 calls in a 2-hour period
is approximately 1 - 0.199 = 0.801, or 80.1%.
Question 5: A stock price follows a geometric Brownian motion with a drift rate of
0.05 and a volatility of 0.2. If the current stock price is $100, what is the expected
stock price after 1 year?

Solution: In a geometric Brownian motion, the stock price at time t follows the stochastic
differential equation:
dS(t) = μ * S(t) * dt + σ * S(t) * dW(t)
Where:
S(t) represents the stock price at time t
μ is the drift rate
σ is the volatility
dt is the differential time interval
dW(t) is the Wiener process
To find the expected stock price after 1 year, we can use the solution of the geometric
Brownian motion:
S(t) = S(0) * exp((μ - 0.5 * σ^2) * t + σ * W(t))
Given that the current stock price S(0) is $100, the drift rate μ is 0.05, and the volatility σ is
0.2, we can calculate the expected stock price after 1 year:
S(1) = S(0) * exp((μ - 0.5 * σ^2) * t + σ * W(t)) = $100 * exp((0.05 - 0.5 * 0.2^2) * 1 + 0.2
* W(1)) For any help regarding Statistics Assignment Help

Since W(1) follows a standard normal distribution, we need to generate a random number
from the standard normal distribution to substitute for W(1). Let's assume we generate a
random number of 0.8.
S(1) = $100 * exp((0.05 - 0.5 * 0.2^2) * 1 + 0.2 * 0.8) = $100 * exp(0.0456)
Calculating this, we get:
S(1) ≈ $104.616
Therefore, the expected stock price after 1 year is approximately $104.616.

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