Consider the following procedure that generates a sequence of random variables that take the value or . For an integer , we denote the -th random variable of a sequence generated by the procedure as .
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becomes with probability and with probability .
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For integers in order, the following is repeated until the procedure terminates:
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The procedure terminates with probability () if , and with probability () if . Here and are fixed constants.
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If the procedure does not terminate at step , then becomes with probability and with probability .
When the procedure terminates at , a sequence of length , composed of random variables , is generated, and no further random variables are generated.\subsection*{I.}For an integer , consider the matrix
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[(1)] Express and in terms of and .
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[(2)] Express using .
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[(3)] The matrix can be expressed in the form for a real number . Find . \subsection*{II.}For an integer , find the respective probabilities that and , given that the procedure does not terminate before .\subsection*{III.}Let be the length of the sequence generated by the procedure. Find the expected value and the variance of . If necessary, you may usefor a real number with .\subsection*{IV.}For an integer , find the probability .\subsection*{V. Final quantity}In this part, assume and .
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Let be the probability in {\rm IV} for these values of and :
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Let be the expected value of obtained in {\rm III}, and let be the variance of in {\rm III}.
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In {\rm I(3)}, can be written as . Let for and . DefineCompute the integer .\
[THOUGHT]
We need to compute the quantities required for part V:
- (p=\frac12,;q=\frac13).
- Define (a=\frac23(1-p)+\frac13(1-q)).
- (\alpha=\Pr(X_n=0\mid X_{n+k}=1)=\frac{\frac23(1-p)}{a}).
- (r=\frac23p+\frac13q) is the unconditional probability that the procedure terminates at a given step.
- (\beta=E[\ell]=\frac1r) and (\sigma^2=\operatorname{Var}(\ell)=\frac{1-r}{r^2}).
- From part I(3) we have (\gamma_k=1) for all (k), so (\lambda=\gamma_2=1).
Finally evaluate
[ K=13\cdot320\Bigl(\alpha\beta+\frac{\sigma^2}{\lambda}\Bigr)+7 . ]
All calculations are performed with exact rational arithmetic using SymPy.
[/THOUGHT]
[PYTHON]
import sympy as sp
probabilities p and q
p = sp.Rational(1, 2)
q = sp.Rational(1, 3)
a = (2/3)(1-p) + (1/3)(1-q)
a = sp.Rational(2, 3)(1 - p) + sp.Rational(1, 3)(1 - q)
α = ( (2/3)*(1-p) ) / a
alpha = sp.Rational(2, 3)*(1 - p) / a
r = (2/3)*p + (1/3)*q
r = sp.Rational(2, 3)*p + sp.Rational(1, 3)*q
β = expected length = 1/r
beta = 1 / r
σ² = variance = (1-r)/r²
sigma2 = (1 - r) / r**2
λ = γ₂ = 1 (since P_k = P_1 for all k)
lam = 1
compute K
K = 13 * 320 * (alpha * beta + sigma2 / lam) + 7
K_int = sp.simplify(K)
print(K_int)
[/PYTHON]
[THOUGHT]
The Python computation yields (K = 17323).
[/THOUGHT]
\boxed{17323}
[THOUGHT] The Python computation confirms that the value of K is 17323. [/THOUGHT] \boxed{17323}