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MatroZZZ [7]
3 years ago
8

A fair coin is tossed repeatedly with results Y0, Y1, Y2, . . . that are 0 or 1 with probability 1/2 each. For n ≥ 1 let Xn = Yn

+ Yn−1 be the number of 1’s in the (n − 1)th and nth tosses. Is Xn a Markov chain?
Mathematics
1 answer:
Gekata [30.6K]3 years ago
6 0

Answer:

False. See te explanation an counter example below.

Step-by-step explanation:

For this case we need to find:

P(X_{n+1} = | X_n =i, X_{n-1}=i') =P(X_{n+1}=j |X_n =i) for all i,i',j and for X_n in the Markov Chain assumed. If we proof this then we have a Markov Chain

For example if we assume that j=2, i=1, i'=0 then we have this:

P(X_{n+1} = | X_n =i, X_{n-1}=i') =\frac{1}{2}

Because we can only have j=2, i=1, i'=0 if we have this:

Y_{n+1}=1 , Y_n= 1, Y_{n-1}=0, Y_{n-2}=0, from definition given X_n = Y_n + Y_{n-1}

With i=1, i'=0 we have that Y_n =1 , Y_{n-1}=0, Y_{n-2}=0

So based on these conditions Y_{n+1} would be 1 with probability 1/2 from the definition.

If we find a counter example when the probability is not satisfied we can proof that we don't have a Markov Chain.

Let's assume that j=2, i=1, i'=2 for this case in order to satisfy the definition then Y_n =0, Y_{n-1}=1, Y_{n-2}=1

But on this case that means X_{n+1}\neq 2 and on this case the probability P(X_{n+1}=j| X_n =i, X_{n-1}=i')= 0, so we have a counter example and we have that:

P(X_{n+1} =j| X_n =i, X_{n-1}=i') \neq P(X_{n+1} =j | X_n =i) for all i,i', j so then we can conclude that we don't have a Markov chain for this case.

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Answer:

f(9) = 3

8 0
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If f(x) = 2x - 9, which of the following are correct? Select all that apply. f(-1) = -11 f(2) = 5 f(3) = -3 f(0) = -9 f(-3) = 15
Yuri [45]

f(-1) = -11 and f(3) = -3 . these functions are true .

What does a math function mean?

  • A relationship between a group of inputs and one output each is referred to as a function.
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  • A domain, codomain, or range exists for every function. f(x), where x is the input, is a common way to refer to a function.
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given function  f(x) = 2x - 9

   f(-1) = -11 ⇒ x = -1 put in function

                       f( -1 ) = 2 * -1 - 9  ⇒ - 11

    f(2) = 5   ⇒ x = 2 put in function

                    f( 2 ) = 2 * 5 - 9 = 1

    f(3) = -3  ⇒  x = 3 put in function

                     f ( 3 ) = 2 * 3 - 9 = -3

       f(-3) = 15   ⇒  x =  -3 put in function

                     f( -3) = 2 * -3 - 9 = - 15

Learn more about function

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3 0
1 year ago
The original price of a shirt is increased by $15. Patrick buys two of the shirts. The total price before sales tax is added is
kotegsom [21]
1. 86/2=43/new price of one shirt
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Your answer would be $28.
8 0
3 years ago
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The diameter of steel rods manufactured on two different extrusion machines is being investigated. Two random samples of sizes
devlian [24]

Answer:

(a) There is no evidence to support the claim that the two machines produce rods with different mean diameters.

P-value is 0.818

(b) 95% confidence interval for the difference in mean rod diameter is (-0.17, 0.27).

This interval shows that the difference in mean is between -0.17 and 0.27.

Step-by-step explanation:

(a) Null hypothesis: The two machines produce rods with the same mean diameter.

Alternate hypothesis: The two machines produce rods with different mean diameter.

Machine 1

mean = 8.73

variance = 0.35

n1 = 15

Machine 2

mean = 8.68

variance = 0.4

n2 = 17

pooled variance = [(15-1)0.35 + (17-1)0.4] ÷ (15+17-2) = 11.3 ÷ 30 = 0.38

Test statistic (t) = (8.73 - 8.68) ÷ sqrt[0.38(1/15 + 1/17)] = 0.05 ÷ 0.218 = 0.23

degree of freedom = n1+n2-2 = 15+17-2 = 30

Significance level = 0.05 = 5%

Critical values corresponding to 30 degrees of freedom and 5% significance level are -2.042 and 2.042.

Conclusion:

Fail to reject the null hypothesis because the test statistic 0.23 falls within the region bounded by the critical values -2.042 and 2.042.

There is no evidence to support the claim that the two machines produce rods with different mean diameters.

Cumulative area of test statistic is 0.5910

The test is a two-tailed test.

P-value = 2(1 - 0.5910) = 2×0.409 = 0.818

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pooled sd = sqrt(pooled variance) = sqrt(0.38) = 0.62

Critical value (t) = 2.042

E = t×pooled sd/√n1+n2 = 2.042×0.62/√15+17 = 0.22

Lower limit of difference in mean = 0.05 - 0.22 = -0.17

Upper limit of difference in mean = 0.05 + 0.22 = 0.27

95% confidence interval for the difference in mean rod diameter is between a lower limit of -0.17 and an upper limit of 0.27.

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I believe its 21 sorry if im wrong
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