answer:Part (a): 1. **Define the problem:** We are given two independent Poisson random variables (X) and (Y) with parameters (lambda > 0) and (mu > 0), respectively. We need to show that the sum (X + Y) is a Poisson random variable with parameter (lambda + mu). 2. **Begin with the probability mass function of a Poisson random variable:** For a Poisson random variable (Z) with parameter (theta), the probability mass function is given by: [ mathrm{P}(Z = k) = frac{theta^k e^{-theta}}{k!}, quad text{for} quad k = 0, 1, 2, ldots ] 3. **Express the probability of (X + Y = n):** [ mathrm{P}(X + Y = n) = sum_{k=0}^{n} mathrm{P}(X = k, Y = n - k) ] 4. **Use the independence property:** [ mathrm{P}(X + Y = n) = sum_{k=0}^{n} mathrm{P}(X = k) mathrm{P}(Y = n - k) ] 5. **Substitute the Poisson probabilities:** [ begin{aligned} mathrm{P}(X + Y = n) &= sum_{k=0}^{n} frac{lambda^k e^{-lambda}}{k!} cdot frac{mu^{n-k} e^{-mu}}{(n-k)!} &= e^{-(lambda + mu)} sum_{k=0}^{n} frac{lambda^k}{k!} cdot frac{mu^{n-k}}{(n-k)!} end{aligned} ] 6. **Recognize the binomial expansion:** [ begin{aligned} mathrm{P}(X + Y = n) &= e^{-(lambda + mu)} cdot frac{1}{n!} sum_{k=0}^{n} frac{n!}{k!(n-k)!} lambda^k mu^{n-k} &= e^{-(lambda + mu)} cdot frac{1}{n!} (lambda + mu)^n &= frac{(lambda + mu)^n e^{-(lambda + mu)}}{n!} end{aligned} ] 7. **Conclude:** Therefore, (X + Y) follows a Poisson distribution with parameter (lambda + mu). [ boxed{ text{Poisson with parameter } lambda + mu } ] Part (b): 1. **Define the conditional probability:** We need to find (mathrm{P}(X = k mid X + Y = n)). 2. **Express the conditional probability:** [ mathrm{P}(X = k mid X + Y = n) = frac{mathrm{P}(X = k, Y = n - k)}{mathrm{P}(X + Y = n)} ] 3. **Substitute the joint and marginal probabilities:** [ mathrm{P}(X = k, Y = n - k) = frac{lambda^k e^{-lambda}}{k!} cdot frac{mu^{n-k} e^{-mu}}{(n-k)!} ] 4. **Use the result from part (a) for (mathrm{P}(X + Y = n)):** [ mathrm{P}(X + Y = n) = frac{(lambda + mu)^n e^{-(lambda + mu)}}{n!} ] 5. **Combine the results:** [ begin{aligned} mathrm{P}(X = k mid X + Y = n) &= frac{frac{lambda^k e^{-lambda}}{k!} cdot frac{mu^{n-k} e^{-mu}}{(n-k)!}}{frac{(lambda + mu)^n e^{-(lambda + mu)}}{n!}} &= frac{lambda^k mu^{n-k} e^{-lambda - mu} n!}{k!(n-k)! (lambda + mu)^n e^{-lambda - mu}} &= frac{n!}{k!(n-k)!} left(frac{lambda}{lambda + mu}right)^k left(frac{mu}{lambda + mu}right)^{n-k} end{aligned} ] 6. **Recognize the binomial distribution:** This is the binomial probability with parameters (n) and (frac{lambda}{lambda + mu}). [ boxed{ text{Binomial with parameters } n text{ and } frac{lambda}{lambda + mu} } ] Part (c): 1. **Identify the task:** We need to find the probability distribution of the difference (X - Y). 2. **Recall characteristics of independent Poisson random variables:** The generating function of a Poisson random variable (Z) with parameter (theta) is given by: [ G_Z(t) = mathbb{E}[t^Z] = e^{theta(t-1)} ] 3. **Construct the generating function for (X - Y):** Since (X) and (Y) are independent, the generating function of (X-Y) is: [ G_{X-Y}(t) = G_X(t) G_Y(t^{-1}) = e^{lambda(t-1)} e^{mu(t^{-1} - 1)} ] 4. **Combine and simplify:** [ G_{X-Y}(t) = e^{lambda(t-1) + mu(t^{-1} - 1)} = e^{-(lambda + mu)} e^{lambda t} e^{mu t^{-1}} ] 5. **Identify as modified Bessel function:** Recognize that this is related to the series expansion of the modified Bessel function (I_k(x)): [ I_k(2x) = x^k sum_{r=0}^{infty} frac{x^{2r}}{r!Gamma(k+r+1)} ] 6. **Conclude:** Therefore, the probability distribution of (X-Y) is given by: [ mathrm{P}(X - Y = k) = e^{-(lambda + mu)} left(frac{lambda}{mu}right)^{k/2} I_k(2sqrt{lambda mu}), quad k = 0, pm 1, pm 2, ldots ] [ boxed{ e^{-(lambda + mu)} left(frac{lambda}{mu}right)^{k/2} I_k(2sqrt{lambda mu}), quad k = 0, pm 1, pm 2, ldots } ]

answer:Given that S_n = n^2 - 2, we can calculate the first term by setting n = 1. Therefore, S_1 = 1^2 - 2 = -1. Since S_n represents the sum of the first n terms of the sequence, the n-th term can be found by subtracting the sum of the first n-1 terms from the sum of the first n terms: a_n = S_n - S_{n-1}. By substituting S_n = n^2 - 2 and S_{n-1} = (n-1)^2 - 2 into the above formula, we get a_n = (n^2 - 2) - [(n-1)^2 - 2]. Expanding the square and simplifying, we obtain begin{align*} a_n &= n^2 - 2 - (n^2 - 2n + 1) &= n^2 - 2 - n^2 + 2n - 1 &= 2n - 1. end{align*} However, this formula does not hold for n=1 as we initially found a_1 = -1. Therefore, the general term is a_n = boxed{begin{cases} -1, & n = 1 2n - 1, & n geq 2 end{cases}}.

answer:Milord, I am Pipo the Fool, a humble jester at thy service. 'Tis an honor to bring merriment and jest to thy esteemed banquet. Mayhap a tune on my lute, a jest, or a trick to astound thy noble guests?

answer:(with a deep bow) Ah, thank you, Your Majesties. 'Tis a grand evening indeed, and I, Pipo the Fool, shall do my utmost to bring laughter and delight to this noble gathering. (producing a small, shiny object from my sleeve) Behold, I present to thee... the Vanishing Coin of Mirth! (holding up the coin and juggling it with a flourish) (winking at the queen) I shall make this coin disappear before thy very eyes, and the eyes of thy noble guests. (miming a series of elaborate gestures, I pretend to drop the coin, only to reveal it has vanished) Ah, but where did it go? Perhaps... (tapping the king's shoulder) 'tis in Your Majesty's ear? (producing the coin from behind the king's ear, to the amusement of the guests) And now, a jest for thee, good sirs and madams. Why did the knight's chicken go to the doctor? (pausing for comedic effect) For it had... (dramatic flourish) fowl breath! (laughter erupts from the guests) (perceiving the queen's smile) Ah, I see Her Majesty enjoys a good pun. Mayhap a song, my queen? A tune on my lute to stir the heart and lift the spirits? (producing a lute from behind the table and beginning to strum a lively melody)