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1. Let X\boldsymbol{X} be N(μ,Σ)N(\boldsymbol{\mu}, \boldsymbol{\Sigma}) with $\bo...
Sep 24, 2024
Solution by Steps
step 1
To find the distribution of 3X1−2X2+X33 X_{1}-2 X_{2}+X_{3}, we first note that if X∼N(μ,Σ)X \sim N(\mu, \Sigma), then any linear combination of the components of XX is also normally distributed. The mean of the linear combination can be calculated as:
\[ E[3X_1 - 2X_2 + X_3] = 3\mu_1 - 2\mu_2 + \mu_3 \]
step 2
Substituting the values of μ1\mu_1, μ2\mu_2, and μ3\mu_3 from μ=(2,−3,1)\boldsymbol{\mu} = (2, -3, 1), we get:
\[ E[3X_1 - 2X_2 + X_3] = 3(2) - 2(-3) + 1 = 6 + 6 + 1 = 13 \]
step 3
Next, we calculate the variance of the linear combination using the covariance matrix Σ\Sigma:
\[ \text{Var}(3X_1 - 2X_2 + X_3) = \begin{pmatrix} 3 & -2 & 1 \end{pmatrix} \Sigma \begin{pmatrix} 3 \\ -2 \\ 1 \end{pmatrix} \]
step 4
Calculating this gives us:
\[ \text{Var}(3X_1 - 2X_2 + X_3) = \begin{pmatrix} 3 & -2 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 & 1 \\ 1 & 3 & 2 \\ 1 & 2 & 2 \end{pmatrix} \begin{pmatrix} 3 \\ -2 \\ 1 \end{pmatrix} \]
step 5
Performing the matrix multiplication results in:
\[ = 3(1)(3) + 3(-2)(1) + 3(1)(1) + (-2)(1)(3) + (-2)(3)(-2) + (-2)(2)(1) + (1)(1)(3) + (1)(2)(-2) + (1)(2)(1) \]
step 6
After calculating, we find:
\[ \text{Var}(3X_1 - 2X_2 + X_3) = 9 + 6 + 3 + 6 + 12 + 4 + 3 - 4 + 2 = 37 \]
step 7
Therefore, the distribution of 3X1−2X2+X33X_1 - 2X_2 + X_3 is:
\[ 3X_1 - 2X_2 + X_3 \sim N(13, 37) \]
step 8
For part (b), we need to find a 2×12 \times 1 vector a\boldsymbol{a} such that X2X_2 and aT(X1X3)\boldsymbol{a}^{T}\binom{X_1}{X_3} are independent. This can be achieved if the covariance between X2X_2 and the linear combination is zero
step 9
The covariance can be expressed as:
\[ \text{Cov}(X_2, \boldsymbol{a}^{T}\binom{X_1}{X_3}) = \boldsymbol{a}^{T} \begin{pmatrix} \text{Cov}(X_2, X_1) \\ \text{Cov}(X_2, X_3) \end{pmatrix} \]
step 10
Setting this equal to zero gives us the equations to solve for a\boldsymbol{a}. From the covariance matrix Σ\Sigma, we find:
\[ \text{Cov}(X_2, X_1) = 1, \quad \text{Cov}(X_2, X_3) = 2 \]
step 11
Thus, we need to solve:
\[ a_1(1) + a_2(2) = 0 \]
step 12
A possible solution is to set a1=2a_1 = 2 and a2=−1a_2 = -1, giving us:
\[ \boldsymbol{a} = \begin{pmatrix} 2 \\ -1 \end{pmatrix} \]
Answer
(a) 3X1−2X2+X3∼N(13,37)3X_1 - 2X_2 + X_3 \sim N(13, 37); (b) a=(2−1)\boldsymbol{a} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}
Key Concept
Linear combinations of normally distributed variables are also normally distributed. Independence can be achieved through zero covariance.
Explanation
The mean and variance of the linear combination were calculated using properties of the normal distribution, and the independence condition was derived from the covariance structure.
Solution by Steps
step 1
To find the expected value of YtY_t, we use the linearity of expectation:
E[Y_t] = E[Y_0 + e_t + e_{t-1} + \cdots + e_1] = E[Y_0] + E[e_t] + E[e_{t-1}] + \cdots + E[e_1]
step 2
Since Y0Y_0 has mean μ0\mu_0 and et,et−1,…,e1e_t, e_{t-1}, \ldots, e_1 are independent with mean 0, we have:
E[Y_t] = \mu_0 + 0 + 0 + \cdots + 0 = \mu_0
step 3
Therefore, we conclude that:
E[Y_t] = \mu_0 \text{ for all } t.
step 4
Now, to find the variance of YtY_t, we use the property of variance for independent random variables:
\operatorname{Var}(Y_t) = \operatorname{Var}(Y_0) + \operatorname{Var}(e_t) + \operatorname{Var}(e_{t-1}) + \cdots + \operatorname{Var}(e_1)
step 5
Since Var⁡(Y0)=σ02\operatorname{Var}(Y_0) = \sigma_0^2 and Var⁡(ei)=σe2\operatorname{Var}(e_i) = \sigma_e^2 for each ii, we have:
\operatorname{Var}(Y_t) = \sigma_0^2 + t \sigma_e^2
step 6
Thus, we conclude that:
\operatorname{Var}(Y_t) = t \sigma_e^2 + \sigma_0^2.
step 7
To find the covariance Cov⁡(Yt,Ys)\operatorname{Cov}(Y_t, Y_s), we consider the minimum of tt and ss:
\operatorname{Cov}(Y_t, Y_s) = \operatorname{Cov}(Y_0, Y_0) + \operatorname{Cov}(e_t, e_t) + \cdots + \operatorname{Cov}(e_{\min(t,s)}, e_{\min(t,s)})
step 8
Since Y0Y_0 contributes σ02\sigma_0^2 and the covariance of eie_i terms is 0 for i≠ji \neq j, we have:
\operatorname{Cov}(Y_t, Y_s) = \sigma_0^2 + \min(t, s) \sigma_e^2
step 9
Therefore, we conclude that:
\operatorname{Cov}(Y_t, Y_s) = \min(t, s) \sigma_e^2 + \sigma_0^2.
step 10
To find the correlation Corr⁡(Yt,Ys)\operatorname{Corr}(Y_t, Y_s), we use the formula:
\operatorname{Corr}(Y_t, Y_s) = \frac{\operatorname{Cov}(Y_t, Y_s)}{\sqrt{\operatorname{Var}(Y_t) \operatorname{Var}(Y_s)}}
step 11
Substituting the covariance and variances we found earlier:
\operatorname{Corr}(Y_t, Y_s) = \frac{\min(t, s) \sigma_e^2 + \sigma_0^2}{\sqrt{(t \sigma_e^2 + \sigma_0^2)(s \sigma_e^2 + \sigma_0^2)}}
step 12
As t→∞t \rightarrow \infty, the correlation approaches 1, thus:
\operatorname{Corr}(Y_t, Y_s) \rightarrow 1 \text{ as } t \rightarrow \infty.
Answer
(a) E[Yt]=μ0E[Y_t] = \mu_0; (b) Var⁡(Yt)=tσe2+σ02\operatorname{Var}(Y_t) = t \sigma_e^2 + \sigma_0^2; (c) Cov⁡(Yt,Ys)=min⁡(t,s)σe2+σ02\operatorname{Cov}(Y_t, Y_s) = \min(t, s) \sigma_e^2 + \sigma_0^2; (d) Corr⁡(Yt,Ys)=tσe2+σ02sσe2+σ02\operatorname{Corr}(Y_t, Y_s) = \sqrt{\frac{t \sigma_e^2 + \sigma_0^2}{s \sigma_e^2 + \sigma_0^2}}
Key Concept
The properties of expectation, variance, covariance, and correlation in random walks and independent random variables.
Explanation
The calculations show how the expected value remains constant, while variance and covariance depend on the time steps, leading to a correlation that approaches 1 as time increases.
Solution by Steps
step 1
To find the expected value of YtY_t, we use the linearity of expectation:
E[Y_t] = E[Y_0 + e_t + e_{t-1} + \cdots + e_1] = E[Y_0] + E[e_t] + E[e_{t-1}] + \cdots + E[e_1]
step 2
Since Y0Y_0 has mean μ0\mu_0 and et,et−1,…,e1e_t, e_{t-1}, \ldots, e_1 are independent with mean 0, we have:
E[Y_t] = \mu_0 + 0 + 0 + \cdots + 0 = \mu_0
step 3
Therefore, we conclude that:
E[Y_t] = \mu_0 \text{ for all } t.
step 4
To find the variance of YtY_t, we use the properties of variance:
\operatorname{Var}(Y_t) = \operatorname{Var}(Y_0 + e_t + e_{t-1} + \cdots + e_1) = \operatorname{Var}(Y_0) + \operatorname{Var}(e_t) + \operatorname{Var}(e_{t-1}) + \cdots + \operatorname{Var}(e_1)
step 5
Since Var⁡(Y0)=σ02\operatorname{Var}(Y_0) = \sigma_0^2 and Var⁡(ei)=σe2\operatorname{Var}(e_i) = \sigma_e^2 for each ii, we have:
\operatorname{Var}(Y_t) = \sigma_0^2 + t \sigma_e^2
step 6
Thus, we conclude that:
\operatorname{Var}(Y_t) = t \sigma_e^2 + \sigma_0^2.
step 7
To find the covariance Cov⁡(Yt,Ys)\operatorname{Cov}(Y_t, Y_s), we note that:
\operatorname{Cov}(Y_t, Y_s) = \operatorname{Cov}(Y_0 + e_t + \cdots + e_1, Y_0 + e_s + \cdots + e_1)
step 8
Since Y0Y_0 contributes σ02\sigma_0^2 and the covariances of the eie_i terms depend on the minimum of tt and ss, we have:
\operatorname{Cov}(Y_t, Y_s) = \min(t, s) \sigma_e^2 + \sigma_0^2.
step 9
Therefore, we conclude that:
\operatorname{Cov}(Y_t, Y_s) = \min(t, s) \sigma_e^2 + \sigma_0^2.
step 10
To find the correlation Corr⁡(Yt,Ys)\operatorname{Corr}(Y_t, Y_s), we use the formula:
\operatorname{Corr}(Y_t, Y_s) = \frac{\operatorname{Cov}(Y_t, Y_s)}{\sqrt{\operatorname{Var}(Y_t) \operatorname{Var}(Y_s)}}
step 11
Substituting the expressions for covariance and variance, we get:
\operatorname{Corr}(Y_t, Y_s) = \frac{\min(t, s) \sigma_e^2 + \sigma_0^2}{\sqrt{(t \sigma_e^2 + \sigma_0^2)(s \sigma_e^2 + \sigma_0^2)}}
step 12
Thus, we conclude that:
\operatorname{Corr}(Y_t, Y_s) = \sqrt{\frac{t \sigma_e^2 + \sigma_0^2}{s \sigma_e^2 + \sigma_0^2}} \text{ for } 0 \leq t \leq s.
Answer
E[Y_t] = \mu_0, \operatorname{Var}(Y_t) = t \sigma_e^2 + \sigma_0^2, \operatorname{Cov}(Y_t, Y_s) = \min(t, s) \sigma_e^2 + \sigma_0^2, \operatorname{Corr}(Y_t, Y_s) = \sqrt{\frac{t \sigma_e^2 + \sigma_0^2}{s \sigma_e^2 + \sigma_0^2}}
Key Concept
The properties of expectation, variance, covariance, and correlation in random walks.
Explanation
The answers demonstrate how to derive the expected value, variance, covariance, and correlation for a random walk with independent increments. Each property follows from the definitions and independence of the components involved.
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