State whether the following statements are true or false. Justify your answer with a short proof or a counter example:
i) Although in one-step transition probability matrix, P, of a Markov chain the sum of each row must necessarily be unity but in higher order transition probability matrices, P(j); j = 2,3,4,...... This rule is not necessary.
ii) For the joint pdf of random variables (X,Y) given by:
iii) Let the r.v. X follows the negative exponential distribution with parameter . Then
iv) One of the examples of non-Poisson queuing system is the (M/G/1): (∞(F1F0) queuing model.
v) If X1,X2,..., X, be a random sample from Np, (μ, Σ), then maximum likelihood estimators of µ and ∑are:
Let's analyze each statement one by one: --- ### i) Although in one-step transition probability matrix, \( P \), of a Markov chain the sum of each row must necessarily be unity, but in higher-order transition probability matrices, \( P^{(j)} \); \( j = 2,3,4,\dots \), this rule is not necessary. Answer: False. Justification: The sum of each row in a transition probability matrix (whether one-step or higher-order) must always be unity. This is because the rows represent the probabilities of transitioning from a given state to all possible states, and these probabilities must sum to 1. For higher-order transition matrices \( _____ ___ _______ __________ ______.
_______ _______ _____ _________ __________ ___ _____ ________ _________ ____ ____.
_____ ________ ___ ___ _________ __________ __________ ___.
_______ _______ _______ ______ _____ _______ ___.
_____ __________ _______ ____ ________ _____ ___ _______ ___ ______ _______.
______ _____ __________ ______ ___ ______ ___.
_____ _______ _____ ________ __________ ________ _______.
___ _______ _______ ___ _____ __________ ______ _______ _____ ___ ______.
_____ __________ _________ _________ _______ ____ _________ _________.
____ _____ ______ ___ ___.
_____ _______ _________ _______ ______ ____ __________.
_________ _____ ________ _________ ___ _______.
_____ __________ _____ ___ _____ ________ ___.
_____ _______ _______ ____ _________ _______ _____ _______.
_________ ____ _____ ______ _________.
_________ __________ _________ __________ ____ _______ ________ ______.
_________ __________ ___ ______ __________ ____.
___ ___ ________ ______ _______ ____ _____ _____ ______.
________ _________ ________ ____ _____.
_______ ___ ______ ___ _______ _______ _____ _____ ___ _____ ___ ______.
_______ ___ ______ ____ _______ ___ ________.
___ ___ _______ ______ __________ __________ _______ ________ ____ ___.
_______ _______ __________ ____ _________ _____ _________ _________ _________ _______ _________ ________.
_________ _____ ___ _____ ____ _______ ______ _________.
_________ __________ ____ _____ ____ _________ __________ ___ ______ ___.
____ ___ ______ _______ ___ ____ ___ _____ ____.
_________ __________ ______ ___ __________ ____ _____ ____ ___ ___ _________.
_____ ____ _____ __________ ___ _________ _______ ____ ___ _________ ______.
____ ________ __________ _________ _______ ______ _______ __________.
_____ ___ ________ __________ _________ _____ ________ __________ ___.
_________ _______ _____ ________ _______ ____ _______ ________ ______ ___.
_____ _______ _________ _____ ___ _________ ____ ______ __________.
_________ _________ _______ __________ ______ ____ _____ ___ _______ ________ ___ _______.
________ __________ ____ ________ _______ ______ __________ ________ ________ ___ _________ __________.
_______ _________ ___ __________ ____ ________ ___ _____ ____ ___ __________.
____ _______ ________ __________ ______ _______ _____ _____ ___ _____.
____ ______ ____ _______ ______ _________.
______ __________ ____ ___ _________ ____ ______ __________ ____ _____ ______.
_____ ____ ___ ___ _______.
___ ______ _________ __________ __________ _______.
____ _________ __________ __________ _________ ________ ________ _______ _______ ___ _____ _______.
_________ ____ ___ __________ ___.
___ ____ ____ ____ _________.
__________ ___ __________ __________ ___ ________ ___ ________ ______ ________ _______.
___ __________ ____ ___ ______ _____ ___ _________ _______ _______ ______ ____.
___ __________ ____ ____ ____ ______ ________ ____ _______ ________ _________.
_______ _____ _______ ___ __________ _________ __________ _____ __________ _____ _________.
________ ____ ____ ________ ________ ___ ___ _______ ____.
_____ _________ __________ ____ ________ ____.
____ ________ _______ ___ __________ ________ ______ _____ _________.
_________ ____ __________ ________ ________ __________ ___ ______ ___ ____.
_______ _______ __________ ____ ___ _________ ________ ______.
_______ ____ _____ __________ ______ ______.
______ __________ ______ _______ __________ _______ ___ _________ _____ _________ ________ ______.
_____ ______ ___ _______ _______ _____.
____ ____ ____ _________ _____ ___ __________.
__________ ______ ________ ______ _______ _________.
______ __________ ____ ___ ______ ______ _________ ________ _______ __________ __________.
______ _________ ________ ___ _________ __________.
_____ _____ _______ ___ _________ ______ _______ ______ ________ ___ _____.
Get Full Answer on WhatsApp
Distinguish between ‘Age Replacement’ and ‘Block Replacement’ policies. Let the lifetimes Y1 ,Y2 ,...., are independently and identically distributed random variables and follows negative exponential distribution with parameter 5. If lifetimes T > 0 and age replacement policy is to be employed, then:
i) find the mean renewal time and
ii) find the long-run average cost per unit time, given the costs C 4 1 = and C 6 2 = units of money.
Consider the following system considering of two switches I and II between two points A and B:
A signal is sent from the point A to point B and is received at B if both the switches I and II are closed. It is assumed that the probabilities of I and II being closed are 0.8 and 0.6 respectively and that P [II is closed | I is closed] = P [II is closed].
Find:
i) The probability that signal is received at B.
ii) The conditional probability that switch I was open, given that the signal was not received at B,
iii) The conditional probability that switch II was open, given that the signal was not received at B.
In an investigation related to a specific type of scores of men and women aged 65 to 70, the mean verbal and performance scores for 101 subjects were found to be:
The sample covariance matrix of the scores was
In order to test the null hypothesis that observations came from a population with mean
apply a suitable test statistic. You may consider α = 01.0 for the test.
For the (M/M/K) : (∞ / FIF )0 queuing model with arrival rate λ and service rate per service channel µ, obtain the steady-state probability of n customers in the system, P . n Hence or otherwise, find the probability that an arrival has to wait.
In a certain state, 58 landfills are classified according to their concentration of three hazardous chemicals: arsenic, barium and mercury. Suppose that the concentration of each one of the three chemicals is characterized as either high or low. If a landfill is chosen at random from among 58 landfills, given the following configuration:
| Barium | ||||
| High Mercury | Low Mercury | |||
| Arsenic | High | Low | High | Low |
| High | 1 | 3 | 5 | 9 |
| Low | 4 | 8 | 10 | 18 |
What is branching process? Give two real examples of branching process.
If P(s) and Pn(s) respectively is the probability generating function (pgf) of the i.i.d. random variables and the random variables {Xn}, where Xn+1 =
Then show that:
Pn(s) = Pn-1(P(s))
and Pn(s) = Pn-1(P(s)).
A study about a population showed that the mobility of population of a state to a village, town and city is in the following percentages:
| From | To | ||
| Village | Town | City | |
| Village | 60 | 25 | 15 |
| Town | 10 | 70 | 20 |
| City | 10 | 30 | 60 |