Create an in-depth explanation of this paper. Start from the RQ page and go linearly from there. Note that "repetition" means the same number of possibilities reoccurring.---Here's the extracted content from the image: iii) As for the case of with repetition, dp_k / dt (t) = 0.87 (0.88 + 0.07) = 0.83 Note that dp_k / dt (t) ≈ f, which signifies the max probability of repetition or no repetition per each state change. Therefore, the expected value of both probability states is mostly the same. Again, P(X = x₂) = (λ² e^(-λt) (1 - e^(-λt))) / (52! (1 - e^(-λt))²) P(X = x₂) > 0.06 = (P(X=x₁) (1-e^(-14.5(0.001)²))) / (1-e^(-14.5(Δt_0->1)²)) P(X = x₂) > 0.06 = (0.06 (1-e^(-14.5(Δt_0->1)²))) / (1-e^(-0.004)² ) Δt_1->2 < 0.0003 t₂ < 14.50003 This confirms our result for t, as it lies mostly within the bounds. Going back to the continuous process definition, we here discuss some characteristics of it: [Note: we use the assumption of Brownian motion which was used by Kolmogorov, as changes in small time intervals are quite small] i) Firstly, we define the rate of change of probabilities throughout the CTMC: dp_k / dt (t) = Σ_j A_jk(t) p_j(t) where A_jk(t) = [δp_k / δu (t; u)]_u=t We input the maximum bound of transition (without repetition in place of the transition matrix. Thus, we get: dp_k / dt (t) = 0.13 ⋅ Σ_j p_j(t) Say, for the first two states: i) Sample Calculation for P(x₁): P(X = x₁) = 277.78 (1 - e^(-14.5Δt_0->1)²) P(X = x₁) = 277.78 (1 - e^(-14.5(0.004)²)) P(X = x₁) = 0.88 Repeat for P(x₂) to get 0.07. ii) Calculation for change in probability, per each change in state: dp_k / dt (t) = 0.13 (0.88 + 0.07) = 0.12 **Extracted Content:** P(X = x₀) = P(N = n₀) In a Poisson Clock, the probability of repetition of the same number of possibilities must only occur at its initial conditions, as the random walk goes further apart from its initial state. To put it exactly, P(T_t < ∞, X₀ = l) = Σ_(n=1)^t f_u^(n) < [p₀t = (14.5)(0.06) = 0.87] As p₀ is a constant value, the probability is summed over the limit [1, t]. From this, we may infer the value of p₀ to be smaller or equal to 0.06. As the probability will only increase as the random walk goes further and further apart, we confirm that:- p_n > 0.06 Now, as for expected value, E(X), it must lie within the lower bound of 0, as there must come upon a case where expected number of possibilities that may occur are null. As for the upper limit: considering the probability f_u^(n) < 0.87, its counterpart (probability of non-recurrence) must be less than 0.13. To calculate for t again, it should be:- f' = p₀t 0.13 = 0.06t t = 2.16 Calculating for t, we get 2.16. Therefore, we encounter two cases for the expected value:- Across cards, with repetition: 0 < E(X) < 14.5 For a single deck, without repetition: 0 < E(X) < 2.16 Since, P(N = n₁) = (52! ⋅ (1-e^(-λΔt_t->∞))) / (λ^2 ⋅ e^(-λ)) P(X = x₁) = (1 - (1-e^(-λΔt_t->∞))) / P(X = x₀) P(X = x₁) = (1 - (1-e^(-λΔt_t->∞))) / P(X = x₀) P(X = x₁) > 0.06 = 277.78(1 - e^(-14.5Δt_0->1))^2 Δt_0->1 < 0.004 t₁ - t₀ < 0.004 t₁ < 14.5004 To solve for t such that we may get E(x₀), we use the Lambert W function which we try to get into form: - (52! p₀)^(1/52) / 52 = -z e^(-z/52) - 52 W ( - (52! p₀)^(1/52) / 52 ) = z - 52 W ( - (52! p₀)^(1/52) / 52 ) = t To get the maximum value of p₀: p(t) = (t^(52) e^(-t)) / 52! d/dt p(t) = (52 t^(51) e^(-t) - t^(52) e^(-t)) / 52! 0 = (52 t^(51) e^(-t) - t^(52) e^(-t)) / 52! t = 52 Thus, we get a maximum value of p₀: inf{n = 1, X = 52} = E(52) = -52 W ((-52! p₀)^(1/52) / 52) for p₀ <= (52^(52) e^(-52)) / 52! [Note: We introduce a bound for p₀ as we progress through the Markov chain. The initial probability remains 1.] Therefore, the argument in the bracket lies between [-1/e, 0), as per the formulation of Lambert W Function. Furthermore, the solution of W must be positive, so it must lie in [-1, ∞). Hence, the entire term is less than positive infinity. This fulfills the requirements of a transient Markov chain: P(T_t < ∞, X₀ = i) = Σ_(n=1)^t f_ii^(n) < 1 To prove for RHS, we know that p₀ <= (52^(52) e^(-52)) / 52!, which is a constant of approximately 0.06 value. Now, inputting it back into equation: 0.06 = (t^(52) e^(-t)) / 52! This will give us a finite value for |t|, which is to approximation 14.5 at max. Therefore, E(x₀) = 14.5. In addition to this, the probability of repetition must at best be less than the value of p₀ summed over the limit [1, t]. Note that this is due to the property that we had discussed earlier: The following content has been extracted from the image: To simplify, λ = rt where r is the rate of occurrence. Here, r = 1. Mathematical Formula: $p_{n-1} = 1 - \left( \frac{e^{-t}}{52!} + \frac{52!(1-e^{-t})}{52! t e^{-t}} + \frac{t^2 e^{-t}(1-e^{-t})^2}{52!(1-e^{-t})} + ... \right)$ In addition, we also encounter a Markovian property of memory m through the justification we had put forward (Assumption 1): Mathematical Formula: $P(X = x_{k-1} | X_{k-2} = x_{k-2}, X_{k-3} = x_{k-3}, X_{k-4} = x_{k-4}, ..., X_0 = x_0) = P(X_n = x_n | X_{n-m} = x_{n-m})$ for all $n > m$ $= p_{n-1}$ In representation, we also get a CTMC as we observe a continuous-time process here as well: Mathematical Formula: $Y(t) := X_{N(t)}$ So, this represents the discrete probability at the N(t)th draw. For t = 0, it will be $X_{N(0)}$ that will need to be imposed. Note that we consider the following equality: Mathematical Formula: $P(X = x_0) = P(N = n_0)$ Clearly, the Poisson clock models the arrivals at time t. Intuitively, at t = 0, the arrivals at time t, which is when the full deck arrives, whose probability must be that of the initial deck's probability distribution as there are no card draws at that time (both probabilities equal to 1). We may use this term then as we go forward. To consider a functional representation of this, such that we may get to n states:- Mathematical Formula: $p_{ij}^{(n)} = P(X_n = j | X_0 = i) = \frac{P(j \cap 0)}{P(i)} = \frac{52! P(j \cap 0)}{t e^{-t}} = \frac{52!(1-e^{-\lambda t \leftrightarrow t})}{t e^{-t}}$ For the case of repeating values, the formulation is trivial. As we will never return to the initial state, the chain is transient. Here, $T_i$ is the hitting time, or the time taken for the initial value to reoccur: Mathematical Formula: $f_{ii}^{(n)} = P(T_i = n | X_0 = i)$ We may also prove this. Proposition 1: In a game of cards, the same number of possibilities will not always reoccur. Mathematical Formula: Theorem 1: $T_i = inf\{n \ge 1, X_n = i\} \Rightarrow P(X_n = i) = \frac{t e^{-t}}{52!}$ The following content is extracted from the image: **Mathematical Formula:** λ = E(N) = Var(N) **Textual Information:** In addition, we get the probability of p₀: **Mathematical Formula:** p₀ = P(N = n₀) = (λ⁵² e⁻ˡᵃ) / 52! **Textual Information:** Now, we consider the conditional probabilities as the state goes on. Note that the Poisson clock does not necessarily follow conditional independence when considered in two different continuous time intervals. **Mathematical Formulas:** P(N = n₁ | N = n₀) = P(n₁, n₀) / P(N = n₀) = P(n₁ n₀) / ( (λ⁵² e⁻ˡᵃ) / 52! ) = (52! P(n₁ n₀)) / (λ⁵² e⁻ˡᵃ) P(N = n₁ | N = n₀) = (52! P(n₁ n₀)) / (λ⁵² e⁻ˡᵃ) **Textual Information:** We may also introduce an approximation for the intersection term. The justification follows that each card draw can be modelled by an exponential distribution CDF, considering the cards as Poisson point processes. We may, as well, consider the two trials as Bernoulli trials if we consider that an event happens, such that P(N ≥ 1) exists. So, joint probabilities may be approximated to p². **Mathematical Formula:** pᵢⱼ = 1 - e⁻ˡᵃᵗ_ (i→j) **Mathematical Formula:** P(N = n₁) = (52! (1 - e⁻ˡᵃᵗ_ (a→b))²) / (λ⁵² e⁻ˡᵃ) **Textual Information:** Therefore, we arrive at Assumption 2. **Textual Information:** Assumption 2: To model Poisson point processes, an exponential distribution may be used wherein the probabilities follow that they are Bernoulli trials. **Textual Information:** Now, considering it again: **Mathematical Formulas:** P(N = n₂ | N = n₁) = (λ⁵² e⁻ˡᵃ (1 - e⁻ˡᵃᵗ_ (n₁→n₂))²) / (52! (1 - e⁻ˡᵃᵗ_ (n₁→n₂))) P(N = n₂) = (λ⁵² e⁻ˡᵃ (1 - e⁻ˡᵃᵗ_ (n₁→n₂))²) / (52! (1 - e⁻ˡᵃᵗ_ (n₁→n₂))) **Textual Information:** So, if we consider a continuous evolution of time, such that: **Mathematical Formulas:** p_ (n-1) = 1 - (p₀ + p₁ + p₂ + ......) p_ (n-1) = 1 - ( (λ⁵² e⁻ˡᵃ) / 52! + (52! (1 - e⁻ˡᵃᵗ_ (a→b))²) / (λ⁵² e⁻ˡᵃ) + (λ⁵² e⁻ˡᵃ (1 - e⁻ˡᵃᵗ_ (i→j))²) / (52! (1 - e⁻ˡᵃᵗ_ (i→j))) + ... ) Here is the complete and accurate extraction of all content from the image: **RQ** What is the maximum and minimum expected number of possibilities of drawing a card (in a deck of 52 cards) with repetition, and without repetition, under the condition of no replacement? **Problem Description and Explanations** In an unbiased, standard deck of cards, there exists an equal probability of 1 / 52 to draw each card. Among the card categories, such as that of Kings, Hearts or Spades, this probability only increases as the number of possibilities decreases. But, the sum of all probabilities of choosing any card at a time must be unity: p₀ + p₁ + p₂ + ...... + p_{n-1} = 1 This must be strictly true for any stage in the game, albeit the probability is dependent on the probability space. Here, the state of the game will only go away and away from the initial state of the game, in a standard game of cards, as the number of cards (possibilities) decrease, that is, in a treatment where we consider the entire deck at a time. To consider the probability distribution of number of possibilities, this can be generally modeled as a random walk: X_k = Y₀ + Y₁ + Y₂ + ...... + Y_{k-1} Now that we have a probability space to work with, the probability density function becomes loosely defined: p(x) = P(X = x) **Assumption 1** So that independence of card draw times are modelled, the card-drawing process is put under continuous time. Formally, we let a Poisson process N(t) with rate lambda encompass the arrival time of draws. At each arrival, the system transitions to a new state according to the DTMC (X_n) on the set of card labels. The observed continuous-time process is then stated by Y(t) := X_{N(t)} of the N(t)^{th} draw at time t. As well as for a fixed time interval t, there will exist a poisson distribution to model the arrivals times. This simplifies the probability density function: N(t) = (λ^k * e^(-λ)) / k! Now, we assume the DTMC we see later on as independent of this poisson process. In its initial state, P(N = n₀): P(N = n₀) = (λ^52 * e^(-λ)) / 52! Note that λ can be further simplified to: