State whether the following statements are True or False. Justify your answer with a short proof or a counter example:
a) If P\mathrm{P} is a transition matrix of a Markov Chain, then all the rows of lim_(nrarr oo)P^(n)\lim _{\mathrm{n} \rightarrow \infty} \mathrm{P}^{\mathrm{n}} are identical.
Answer:
The statement “If PP is a transition matrix of a Markov Chain, then all the rows of lim_(n rarr oo)P^(n)\lim_{{n \to \infty}} P^n are identical” is generally not true for all Markov Chains. It is true under certain conditions, such as if the Markov Chain is irreducible, aperiodic, and positive recurrent (i.e., it is ergodic).
Counterexample:
Consider a simple Markov Chain with two states AA and BB and the following transition matrix:
In this case, the Markov Chain is not irreducible (it consists of two disconnected states). The limit lim_(n rarr oo)P^(n)\lim_{{n \to \infty}} P^n exists and is:
Here, the covariance between XX and YY is -1-1, which is not positive. Therefore, the statement is false.
Additional Notes:
The diagonal elements of a variance-covariance matrix, which represent variances, are always non-negative because variance cannot be negative.
Off-diagonal elements, which represent covariances, can be negative, zero, or positive, depending on the relationship between the variables involved.
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c) If X_(1),X_(2),X_(3)X_1, X_2, X_3 are iid from N_(2)(mu,Sigma)N_2(\mu, \Sigma), then (X_(1)+X_(2)+X_(3))/(3)\frac{X_1+X_2+X_3}{3} follows N_(2)(mu,(1)/(3)Sigma)N_2\left(\mu, \frac{1}{3} \Sigma\right).
Answer:
The statement “If X_(1),X_(2),X_(3)X_1, X_2, X_3 are iid from N_(2)(mu,Sigma)N_2(\mu, \Sigma), then (X_(1)+X_(2)+X_(3))/(3)\frac{X_1+X_2+X_3}{3} follows N_(2)(mu,(1)/(3)Sigma)N_2\left(\mu, \frac{1}{3} \Sigma\right)” is true.
Justification:
Mean: The mean of (X_(1)+X_(2)+X_(3))/(3)\frac{X_1+X_2+X_3}{3} is (mu+mu+mu)/(3)=mu\frac{\mu + \mu + \mu}{3} = \mu.
Covariance Matrix: The covariance matrix of X_(1)+X_(2)+X_(3)X_1+X_2+X_3 is Sigma+Sigma+Sigma=3Sigma\Sigma + \Sigma + \Sigma = 3\Sigma because X_(1),X_(2),X_(3)X_1, X_2, X_3 are independent. Therefore, the covariance matrix of (X_(1)+X_(2)+X_(3))/(3)\frac{X_1+X_2+X_3}{3} is (1)/(3^(2))(3Sigma)=(1)/(3)Sigma\frac{1}{3^2}(3\Sigma) = \frac{1}{3}\Sigma.
Since both the mean and the covariance matrix match the given distribution N_(2)(mu,(1)/(3)Sigma)N_2\left(\mu, \frac{1}{3} \Sigma\right), the statement is true.
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d) The partial correlation coefficients and multiple correlation coefficients lie between -1 and 1.
Answer:
The statement “The partial correlation coefficients and multiple correlation coefficients lie between -1 and 1” is true.
Justification:
Partial Correlation Coefficients: The partial correlation coefficient measures the strength and direction of the relationship between two variables while controlling for the effect of one or more other variables. Mathematically, it is defined in a way that ensures its value lies between -1 and 1, inclusive. Specifically, it is computed as the correlation between the residuals resulting from the linear regression of each variable against the control variables. Since residuals are uncorrelated with the predicted values, the partial correlation coefficient is constrained to be between -1 and 1.
Multiple Correlation Coefficients: The multiple correlation coefficient RR is defined as the square root of the coefficient of determination R^(2)R^2, which is the proportion of the variance in the dependent variable that is predictable from the independent variables in a multiple regression model. Since R^(2)R^2 is between 0 and 1, RR must be between 0 and 1. However, RR can be negative if the model includes a constant term and the slope is negative, but its absolute value will still be between 0 and 1.
Therefore, both the partial correlation coefficients and multiple correlation coefficients are bounded between -1 and 1, making the statement true.
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e) For a renewal function M_(t),lim_(t rarr0)(M_(t))/(t)=(1)/(mu)M_t, \lim _{t \rightarrow 0} \frac{M_t}{t}=\frac{1}{\mu}.
Answer:
The statement “For a renewal function M_(t),lim_(t rarr0)(M_(t))/(t)=(1)/(mu)M_t, \lim_{{t \rightarrow 0}} \frac{M_t}{t} = \frac{1}{\mu}” is generally true under certain conditions.
Justification:
A renewal function M_(t)M_t is defined as the expected number of renewals (or arrivals, or events) that have occurred by time tt. Mathematically, it is often defined as:
M_(t)=E[N(t)]M_t = \mathbb{E}[N(t)]
where N(t)N(t) is the number of renewals by time tt.
The mean inter-arrival time (or mean time between renewals) is denoted by mu\mu and is defined as:
mu=E[X]\mu = \mathbb{E}[X]
where XX is the random variable representing the time between renewals.
Under the assumption that mu < oo\mu < \infty and the distribution of inter-arrival times has a finite variance, it is generally true in renewal theory that:
This result is often derived using advanced methods in renewal theory and stochastic processes, and it provides a way to relate the renewal function to the mean inter-arrival time mu\mu.
So, the statement is true under the conditions that mu < oo\mu < \infty and the distribution of inter-arrival times has a finite variance.
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Question:-02
a) Let (X,Y)(\mathrm{X}, \mathrm{Y}) have the joint p.d.f. given by:
f(x,y)={[1″,”,” if “|y| < x;0 < x < 1],[0″,”,” otherwise “]:}f(x, y)= \begin{cases}1, & \text { if }|y|<x ; 0<x<1 \\ 0, & \text { otherwise }\end{cases}
i) Find the marginal p.d.f.’s of X\mathrm{X} and Y\mathrm{Y}.
ii) Test the independence of X\mathrm{X} and Y\mathrm{Y}.
iii) Find the conditional distribution of XX given Y=yY=y.
iv) Compute E(X∣Y=y)\mathrm{E}(\mathrm{X} \mid \mathrm{Y}=\mathrm{y}) and E(Y∣X=x)\mathrm{E}(\mathrm{Y} \mid \mathrm{X}=\mathrm{x}).
Answer:
i) Marginal p.d.f.’s of XX and YY
Marginal p.d.f. of XX
f_(X)(x)=int_(-x)^(x)1dy=2x quad”for “0 < x < 1f_X(x) = \int_{-x}^{x} 1 \, dy = 2x \quad \text{for } 0 < x < 1
i) Find the marginal distribution of X\mathrm{X} and Y\mathrm{Y}.
ii) Find the conditional distribution of XX given Y=1Y=1.
iii) Test the independence of variable sX\mathrm{s} X and Y\mathrm{Y}.
iv) Find V[(Y)/(X)=x]V\left[\frac{Y}{X}=x\right].
Answer:
Introduction
In this problem, we are given the joint probability density function (pdf) of two discrete random variables XX and YY. We are tasked with:
Finding the marginal distribution of XX and YY.
Finding the conditional distribution of XX given Y=1Y=1.
Testing the independence of XX and YY.
Finding the variance V[(Y)/(X)=x]V\left[\frac{Y}{X}=x\right].
Let’s proceed to solve each part step-by-step.
Part i: Marginal Distribution of XX and YY
Marginal Distribution of XX
The marginal distribution of XX can be found by summing the probabilities along each row of the table for each value of XX.
The marginal distribution of XX is P(X=2)=0.34P(X=2) = 0.34, P(X=3)=0.33P(X=3) = 0.33, P(X=4)=0.19P(X=4) = 0.19, P(X=5)=0.14P(X=5) = 0.14.
The marginal distribution of YY is P(Y=0)=0.03P(Y=0) = 0.03, P(Y=1)=0.80P(Y=1) = 0.80, P(Y=2)=0.17P(Y=2) = 0.17.
The conditional distribution of XX given Y=1Y=1 is P(X=2|Y=1)=0.425P(X=2|Y=1) = 0.425, P(X=3|Y=1)=0.375P(X=3|Y=1) = 0.375, P(X=4|Y=1)=0.2P(X=4|Y=1) = 0.2, P(X=5|Y=1)=0P(X=5|Y=1) = 0.
XX and YY are not independent.
The variance V[(Y)/(X)=x]V\left[\frac{Y}{X}=x\right] is approximately 0.02870.0287.