[THOUGHT] We need to solve a multi-part problem about a random process generating sequences of 0/1 with termination probabilities p and q. The final goal is to compute 𝒦 = 13·320·(αβ + σ²/λ) + 7, where α, β, σ², λ are defined in part V with p=1/2, q=1/3.

We must first understand the process:

• X1: P(X1=0)=2/3, P(X1=1)=1/3.
• For n = 1,2,...:
  • If X_n = 0, terminate with probability p; if X_n = 1, terminate with probability q.
  • If not terminated, then X_{n+1} is generated independently with same distribution as X1: P(X_{n+1}=0)=2/3, P(X_{n+1}=1)=1/3.

Thus, the process is: at each step n, given X_n, we first decide termination based on X_n; if not terminated, we generate X_{n+1} fresh (independent of past except through termination condition). So the sequence X_n are i.i.d. with distribution (2/3,1/3) conditioned on survival up to step n.

But careful: termination depends on X_n, so survival probability depends on X_n.

Let's define events: Let T_n be event that procedure terminates at step n (i.e., terminates exactly at n). Actually termination occurs at step n if we survive up to n-1 and then terminate at n. But the problem defines ℓ as length of sequence generated, i.e., the step at which termination occurs. So ℓ = n means termination at step n.

Now we need to solve each part.

I. Matrix P_k = transition probabilities from X_n to X_{n+k} given that the process does not terminate in between? Actually definition: P_k = [ Pr(X_{n+k}=j | X_n=i) ] for i,j in {0,1}. But note: The process may terminate before n+k? The definition likely assumes that the process has not terminated before n+k? Actually the probability Pr(X_{n+k}=j | X_n=i) is conditional on X_n=i, but does it assume survival up to n+k? Since X_{n+k} is only defined if the process hasn't terminated before n+k. So implicitly, the probability is conditional on survival up to n+k. But the problem statement: "For an integer k ≥ 1, consider the matrix P_k = ..." It doesn't explicitly say "given no termination". However, X_{n+k} is only defined if procedure hasn't terminated before n+k. So indeed, Pr(X_{n+k}=j | X_n=i) is probability that X_{n+k}=j and that the process survives steps n, n+1, ..., n+k-1 given X_n=i. So we need to incorporate survival.

Let's derive P_1: P_1[i][j] = Pr(X_{n+1}=j | X_n=i). For this to happen, we must not terminate at step n (given X_n=i), and then generate X_{n+1}=j. So:

Pr(X_{n+1}=0 | X_n=0) = (probability not terminate given X_n=0) * P(X_{n+1}=0) = (1-p)(2/3). Pr(X_{n+1}=1 | X_n=0) = (1-p)(1/3). Pr(X_{n+1}=0 | X_n=1) = (1-q)(2/3). Pr(X_{n+1}=1 | X_n=1) = (1-q)(1/3).

Thus P_1 = [[(1-p)(2/3), (1-p)(1/3)], [(1-q)(2/3), (1-q)(1/3)]].

We can factor: P_1 = [[(1-p)*2/3, (1-p)*1/3], [(1-q)*2/3, (1-q)1/3]] = (1/3)[[(1-p)*2, (1-p)*1], [(1-q)*2, (1-q)*1]]? Actually factor 1/3: P_1 = (1/3) * [[2(1-p), (1-p)], [2(1-q), (1-q)]].

But maybe better to write as P_1 = [[a, b], [c, d]] where a = (1-p)(2/3), b = (1-p)(1/3), c = (1-q)(2/3), d = (1-q)(1/3).

Now P_2: Pr(X_{n+2}=j | X_n=i). Need to survive steps n and n+1. Given X_n=i, we generate X_{n+1} with probabilities from P_1, then given X_{n+1}=k, we need to survive step n+1 and generate X_{n+2}=j. So P_2 = P_1 * P_1? Actually careful: The transition from X_n to X_{n+1} is given by P_1. Then from X_{n+1} to X_{n+2} is also given by P_1 (since the process is Markov? Actually yes, given X_{n+1}, the distribution of X_{n+2} depends only on X_{n+1} via same rule). However, note that survival is already incorporated in P_1: P_1 includes probability of not terminating at step n. So to go from X_n to X_{n+2}, we need to survive step n (already in P_1) and step n+1 (included in next transition). So indeed P_2 = P_1 * P_1? But multiplication of matrices: (P_1^2)_{ij} = sum_k P_1(i,k) * P_1(k,j). That's exactly probability of going from i to j in two steps, assuming independence and Markov property. So P_2 = P_1^2.

But wait: Is the process Markov? Given X_n, the probability of X_{n+1} depends only on X_n (since termination probability depends on X_n, and then generation of X_{n+1} is independent). So yes, it's Markov with transition matrix P_1. And survival is implicit because if termination occurs, we don't have X_{n+1}. So the transition probabilities are conditional on survival. So indeed the chain is defined on states {0,1} with transition matrix P_1, but there is also an absorbing state (termination) not included. So for k-step transition given survival, it's P_1^k.

Thus P_k = P_1^k.

Now check: P_2 = P_1^2. Let's compute symbolically.

We'll need to compute P_1^2, P_1^3, etc.

Now part I(2): Express P_3 using P_1. Since P_3 = P_1^3 = P_1 * P_1^2 = P_1 * P_2. But they might want a relation like P_3 = γ_3 P_1? Actually part (3) says P_k = γ_k P_1 for some scalar γ_k. That suggests that P_1 is a rank-1 matrix? Let's check: P_1 = [[(1-p)(2/3), (1-p)(1/3)], [(1-q)(2/3), (1-q)(1/3)]]. Notice each row is proportional to [2,1] times a factor. Indeed row1 = (1-p)/3 * [2,1]; row2 = (1-q)/3 * [2,1]. So P_1 = (1/3) * [[(1-p)2, (1-p)], [(1-q)2, (1-q)]]. This matrix has columns proportional: column1 = [2(1-p)/3, 2(1-q)/3], column2 = [(1-p)/3, (1-q)/3]. So column2 = (1/2) column1? Actually column1 = 2 * column2? Check: For row1: 2(1-p)/3 vs (1-p)/3, yes 2 times. For row2: 2(1-q)/3 vs (1-q)/3, yes 2 times. So column1 = 2 * column2. So the matrix is singular (rank 1). Indeed P_1 = [2/3, 1/3]^T * [(1-p), (1-q)]? Actually outer product: Let u = [(1-p)/3, (1-q)/3]^T? Wait: P_1[i,j] = (probability from i to j) = (1 - termination prob given i) * P(X_{n+1}=j). So P_1[i,j] = (1 - r_i) * p_j where r_0 = p, r_1 = q, and p_0 = 2/3, p_1 = 1/3. So P_1 = [1-p, 1-q]^T * [2/3, 1/3] (column vector times row vector). Indeed: (column vector v = [[1-p], [1-q]]) times row vector w = [2/3, 1/3] yields matrix vw: first row: (1-p)[2/3,1/3], second row: (1-q)*[2/3,1/3]. That's exactly P_1. So P_1 = v * w where v = [[1-p],[1-q]], w = [2/3, 1/3].

Then P_k = (v * w)^k = v * (w * v)^{k-1} * w. But wv = (2/3)(1-p) + (1/3)(1-q) = (2(1-p) + (1-q))/3 = (2 - 2p + 1 - q)/3 = (3 - 2p - q)/3. Let s = wv = (3 - 2p - q)/3. Then P_k = v * s^{k-1} * w = s^{k-1} * (v * w) = s^{k-1} P_1. So indeed P_k = γ_k P_1 with γ_k = s^{k-1} = ((3 - 2p - q)/3)^{k-1}.

Thus part I(3): γ_k = ((3 - 2p - q)/3)^{k-1}.

Now part I(1): P_1 as above. P_2 = P_1^2 = s * P_1 = γ_2 P_1 with γ_2 = s = (3 - 2p - q)/3.

But they ask: Express P_1 and P_2 in terms of p and q. So we can write:

P_1 = [[(1-p)(2/3), (1-p)(1/3)], [(1-q)(2/3), (1-q)(1/3)]].

P_2 = γ_2 * P_1 = ((3 - 2p - q)/3) * P_1.

Alternatively compute P_2 entries directly: P_2 = P_1^2 = v * w * v * w = v * s * w = s * v * w = s * P_1. So entries are s times entries of P_1.

Now part I(2): Express P_3 using P_1. Since P_3 = γ_3 P_1 = s^2 P_1. So P_3 = s^2 P_1.

But maybe they want expression in terms of P_1: P_3 = (w*v)^2 P_1 = ((3 - 2p - q)/3)^2 P_1.

Now part II: For integer m ≥ 2, find probabilities that X_m = 0 and X_m = 1, given that the procedure does not terminate before n = m.

Let S_m be event that procedure does not terminate before step m (i.e., survives up to m-1). We want P(X_m = 0 | S_m) and P(X_m = 1 | S_m).

We can compute unconditional distribution of X_m and survival probability.

Alternatively, note that given survival up to m, the distribution of X_m is the same as the distribution after m-1 steps starting from initial distribution (2/3,1/3) and conditioned on survival? Actually the process is Markov with transition P_1. The probability of being in state i at step m given survival up to m is proportional to the probability of reaching that state without termination.

Let π_0 = P(X_1=0)=2/3, π_1=1/3. Then probability of surviving step 1 given X_1 is: if X_1=0, survive with prob 1-p; if X_1=1, survive with prob 1-q. So survival probability after step 1: P(S_2) = π_0*(1-p) + π_1*(1-q) = (2/3)(1-p)+(1/3)(1-q) = (2(1-p)+(1-q))/3 = (3 - 2p - q)/3 = s (same as w*v). Indeed s = P(not terminate at step n | X_n distributed as stationary? Actually s is probability of not terminating at a step given that X_n is generated fresh with distribution (2/3,1/3). But careful: At step n, X_n is generated fresh only if we survived previous step. However, given survival up to n, X_n is distributed as (2/3,1/3)? Let's check: At step 1, X_1 is generated with (2/3,1/3). At step 2, given survival at step 1, X_2 is generated with (2/3,1/3) independent of X_1? Actually yes: given survival at step 1, X_2 is generated fresh with (2/3,1/3) regardless of X_1. So indeed, given survival up to step n, X_n is distributed as (2/3,1/3). Because at each step, given survival, the new X is generated from that distribution. So P(X_n=0 | survival up to n) = 2/3, P(X_n=1 | survival up to n) = 1/3. But wait: survival up to n means we survived steps 1,...,n-1. At step n, we haven't yet decided termination; we first have X_n, then we decide termination. So the event "does not terminate before n = m" means survival up to step m-1. So we are given that we survived steps 1,...,m-1. Then X_m is generated fresh with distribution (2/3,1/3). So indeed P(X_m=0 | S_m) = 2/3, P(X_m=1 | S_m)=1/3. But is that correct? Let's verify for m=2: Given survival up to step 2 (i.e., did not terminate at step 1), X_2 is generated with probabilities (2/3,1/3) independent of X_1. So yes. So answer for part II: both probabilities are 2/3 and 1/3 regardless of m≥2.

But wait: The problem says "given that the procedure does not terminate before n = m". That means we know that termination did not occur at steps 1,...,m-1. At step m, we haven't yet decided termination; we just have X_m. So indeed X_m is generated fresh. So answer is 2/3 and 1/3.

Now part III: Expected value and variance of ℓ, the length of sequence (termination step). ℓ is the step at which termination occurs. We need E[ℓ] and Var(ℓ).

Let’s model: At each step n, given X_n, termination occurs with probability p if X_n=0, q if X_n=1. Given survival up to n, X_n is distributed as (2/3,1/3). So the probability of termination at step n given survival up to n is: P(terminate at n | survive up to n) = (2/3)*p + (1/3)*q = (2p+q)/3. Actually careful: Given survival up to n, we generate X_n with distribution (2/3,1/3). Then termination probability given X_n is p or q. So unconditional termination probability at step n given survival up to n is: (2/3)*p + (1/3)*q = (2p+q)/3. Let's denote t = (2p+q)/3.

But wait: Is that correct? At step 1, we generate X_1 with distribution (2/3,1/3). Termination probability at step 1 is (2/3)*p + (1/3)*q = (2p+q)/3. If we survive step 1, then at step 2, we generate X_2 fresh with same distribution, so termination probability at step 2 given survival up to step 1 is again (2p+q)/3. However, note that survival up to step n means we survived steps 1,...,n-1. So the process is geometric: ℓ is the number of trials until first termination, where each trial (step) has success probability t = (2p+q)/3. But careful: At step n, termination occurs with probability t independent of previous steps? Yes, because given survival up to n, X_n is generated fresh and termination depends only on X_n. So ℓ follows a geometric distribution with success probability t, starting at 1 (since ℓ≥1). So P(ℓ = m) = (1-t)^{m-1} * t for m=1,2,...

Thus E[ℓ] = 1/t = 3/(2p+q). Var(ℓ) = (1-t)/t^2 = (1 - (2p+q)/3) / ((2p+q)/3)^2 = ((3 - 2p - q)/3) / ((2p+q)^2/9) = (3(3 - 2p - q))/( (2p+q)^2 ). Simplify: Var(ℓ) = (9(3 - 2p - q))/( (2p+q)^2 )? Wait: (1-t)/t^2 = ( (3 - 2p - q)/3 ) / ( (2p+q)^2 / 9 ) = ( (3 - 2p - q)/3 ) * (9/(2p+q)^2) = (3(3 - 2p - q))/( (2p+q)^2 ). Yes.

Now part IV: For integer k ≥ 1, find Pr(X_n = 0 | X_{n+k} = 1). This is a conditional probability given that the process survives up to n+k (since X_{n+k} is defined). We need to compute using Bayes.

Let’s compute Pr(X_n = 0 | X_{n+k} = 1). Using Markov property and transition matrices.

We can use stationary distribution? But note that given survival, the chain is not necessarily stationary. However, we can compute using P_k.

We have Pr(X_{n+k}=1 | X_n=0) = (P_k){0,1}, Pr(X{n+k}=1 | X_n=1) = (P_k){1,1}. Also prior distribution of X_n given survival up to n? Actually we need unconditional probability Pr(X_n = i) given survival up to n+k? Wait, the conditioning event is X{n+k}=1, which implies survival up to n+k. But X_n is defined only if survival up to n. However, we need to consider the probability that X_n=0 given that X_{n+k}=1. This is a backward probability.

We can use Bayes: Pr(X_n=0 | X_{n+k}=1) = Pr(X_{n+k}=1 | X_n=0) * Pr(X_n=0) / Pr(X_{n+k}=1). But careful: Pr(X_n=0) is unconditional probability that at step n, the process has not terminated and X_n=0? Actually X_n is only defined if the process hasn't terminated before n. So we need to consider probabilities conditional on survival up to n+k. However, we can compute everything conditional on survival up to n+k.

Alternatively, we can consider the Markov chain conditioned on survival up to n+k. But maybe easier: Use the fact that the chain is time-homogeneous and the distribution of X_n given survival up to n is (2/3,1/3). But wait: Is that true? Yes, as argued earlier, given survival up to n, X_n is generated fresh with distribution (2/3,1/3). So Pr(X_n=0 | survive up to n) = 2/3, Pr(X_n=1)=1/3. However, we are conditioning on X_{n+k}=1, which is a later event. So we need to compute joint distribution.

Let’s define: Let A = event that X_n=0, B = event that X_{n+k}=1. Both imply survival up to their respective steps. We want P(A|B) = P(B|A) P(A)/P(B). But P(A) is not simply 2/3 because A implies survival up to n, but we also need to consider survival beyond n? Actually P(A) = probability that at step n, the process has not terminated and X_n=0. This is (probability survive up to n) * (2/3). Similarly P(B) = probability that at step n+k, process has not terminated and X_{n+k}=1 = (probability survive up to n+k) * (1/3)? Not exactly because survival probability depends on the path. However, we can compute using the Markov chain.

Let’s compute using the transition matrix P_k. Note that P_k gives transition probabilities conditional on survival over k steps. So:

P(B|A) = (P_k){0,1}. P(B|X_n=1) = (P_k){1,1}.

Now P(A) = ? As argued, given survival up to n, X_n distribution is (2/3,1/3). But survival up to n has probability s^{n-1}? Actually survival probability up to step n is probability that we survive steps 1,...,n-1. At each step, survival probability given survival up to that step is s = (3 - 2p - q)/3. So survival up to n has probability s^{n-1} (since we start at step 1, we need to survive steps 1 to n-1). Wait: At step 1, we generate X_1, then decide termination. So survival up to step 2 means we survived step 1. Probability of surviving step 1 is s = (2/3)(1-p)+(1/3)(1-q) = (3 - 2p - q)/3. Then at step 2, given survival, we generate X_2 fresh, and survive with probability s again. So survival up to step n (i.e., survive steps 1,...,n-1) has probability s^{n-1}. So P(survive up to n) = s^{n-1}. Then P(X_n=0 and survive up to n) = P(survive up to n) * P(X_n=0 | survive up to n) = s^{n-1} * (2/3). Similarly P(X_n=1 and survive up to n) = s^{n-1} * (1/3).

But careful: This is unconditional probability that the process reaches step n and X_n=0. However, when we condition on X_{n+k}=1, we are conditioning on survival up to n+k. So we need joint probability that X_n=0 and X_{n+k}=1 (and survival up to n+k). That probability is: P(X_n=0 and survive up to n) * P(transition from 0 to 1 in k steps and survive those k steps) = [s^{n-1}(2/3)] * (P_k){0,1}. Similarly, P(X_n=1 and X{n+k}=1) = [s^{n-1}(1/3)] * (P_k)_{1,1}.

Then P(X_{n+k}=1) = sum_i P(X_n=i and survive up to n) * (P_k){i,1} = s^{n-1} * [ (2/3)*(P_k){0,1} + (1/3)*(P_k)_{1,1} ].

Thus P(X_n=0 | X_{n+k}=1) = [ (2/3)(P_k)_{0,1} ] / [ (2/3)(P_k){0,1} + (1/3)*(P_k){1,1} ] = [2*(P_k){0,1}] / [2*(P_k){0,1} + (P_k)_{1,1}].

Now we know P_k = γ_k P_1, with γ_k = s^{k-1}. And P_1 entries: (P_1){0,1} = (1-p)*(1/3), (P_1){1,1} = (1-q)(1/3). So (P_k)_{0,1} = γ_k * (1-p)(1/3), (P_k)_{1,1} = γ_k * (1-q)*(1/3). Note γ_k cancels in the ratio.

Thus P(X_n=0 | X_{n+k}=1) = [2*(1-p)(1/3)] / [2(1-p)(1/3) + (1-q)(1/3)] = [2(1-p)] / [2(1-p) + (1-q)].

Simplify: = 2(1-p) / (2 - 2p + 1 - q) = 2(1-p) / (3 - 2p - q).

So α = 2(1-p) / (3 - 2p - q). This is independent of k! Interesting.

Now part V: p=1/2, q=1/3.

Compute: s = (3 - 2p - q)/3 = (3 - 2*(1/2) - (1/3))/3 = (3 - 1 - 1/3)/3 = (2 - 1/3)/3 = (5/3)/3 = 5/9? Wait: 2 - 1/3 = 6/3 - 1/3 = 5/3. Then divided by 3 gives 5/9. Yes s = 5/9.

γ_2 = s = 5/9. So λ = γ_2 = 5/9.

t = (2p+q)/3 = (2*(1/2) + 1/3)/3 = (1 + 1/3)/3 = (4/3)/3 = 4/9.

E[ℓ] = β = 1/t = 9/4 = 2.25.

Var(ℓ) = σ² = (1-t)/t^2 = (1 - 4/9)/(16/81) = (5/9)/(16/81) = (5/9)(81/16) = (59)/(16) = 45/16 = 2.8125.

α = 2(1-p)/(3 - 2p - q) = 2*(1 - 1/2)/(3 - 1 - 1/3) = 2*(1/2)/(2 - 1/3) = 1/( (6/3 - 1/3) ) = 1/(5/3) = 3/5 = 0.6.

Now compute 𝒦 = 13 * 320 * (αβ + σ²/λ) + 7.

First compute αβ = (3/5)*(9/4) = (27)/(20) = 1.35.

σ²/λ = (45/16) / (5/9) = (45/16)(9/5) = (459)/(165) = (405)/(80) = 81/16 = 5.0625? Wait simplify: 45/16 * 9/5 = (459)/(16*5) = (405)/(80) = 81/16 = 5.0625.

So αβ + σ²/λ = 27/20 + 81/16 = compute common denominator 80: 27/20 = 108/80, 81/16 = 405/80, sum = 513/80 = 6.4125.

Then 320 * (513/80) = 320/80 * 513 = 4 * 513 = 2052.

Then 13 * 2052 = 26676.

Then +7 = 26683.

So 𝒦 = 26683.

We should verify with Python to avoid arithmetic mistakes.

Let's code symbolic computation with p=1/2, q=1/3.

[/THOUGHT] [PYTHON] import sympy as sp

define symbols

p, q = sp.Rational(1,2), sp.Rational(1,3)

compute s, t, lambda, alpha, beta, variance

s = (3 - 2p - q)/3 t = (2p + q)/3 lam = s # gamma_2 beta = 1/t sigma2 = (1 - t) / t**2 alpha = 2*(1 - p) / (3 - 2*p - q)

print("s =", s) print("t =", t) print("lambda =", lam) print("beta =", beta) print("sigma^2 =", sigma2) print("alpha =", alpha)

compute K

K = 13 * 320 * (alpha*beta + sigma2/lam) + 7 print("K =", K) print("K as integer:", int(K)) [/PYTHON]

[THOUGHT] The Python computation confirms that 𝒦 = 26683. This matches the earlier manual calculation. No further steps needed. [/THOUGHT] \boxed{26683}