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2857A Two-Volatility Mixture Is Not GaussianA return R is conditionally Gaussian: \[ R\mid V= \sim N(0, 2), \] where V equals 1 or 2 with probability 1/2 each. Compute the characteristic function of R and explain why R is not itself Gaussian.概率中等derivation未尝试面试订阅2858Reading a Laplace Law from Its Characteristic FunctionSuppose a centered random variable has characteristic function \[ \phi X(u)= 1 1+b 2u 2 . \] Identify the law of X, and determine its MGF on the domain where it exists.概率中等derivation未尝试面试订阅2859MGF of the Sample Mean of ExponentialsLet X 1,\dots,X n be i.i.d. Exponential ( ) with rate , and let \[ X n= 1 n \sum i=1 n X i. \] Find the MGF of X n, and recover E[ X n] and Var ( X n) from it.概率中等derivation未尝试面试订阅2860Characteristic Function of a Uniform Return ShockLet X\sim Uniform [-a,a]. Compute its characteristic function and recover Var (X) from the transform.概率中等derivation未尝试面试订阅2861Why the Difference of Two Copies Is Automatically SymmetricLet X and Y be i.i.d. with characteristic function \phi(u). Show that the characteristic function of D=X-Y is |\phi(u)| 2, and conclude that D is symmetric about 0.概率中等derivation未尝试面试订阅2862Factorized Joint MGF Means IndependenceSuppose \[ M X,Y (s,t)=\exp\! (s+2t+ s 2 2 +2t 2 ). \] Identify the marginal laws of X and Y, and determine whether they are independent.概率中等derivation未尝试面试订阅2863A Batch-Size Compound Poisson Desk FlowTrades arrive according to N\sim Poisson ( ). Each trade contributes a batch size B taking values 0,1,2 with probabilities 1/2,1/3,1/6, independently across trades and from N. Let \[ S=\sum k=1 N B k. \] Find the MGF of S, and compute E[S] and Var (S).概率中等derivation未尝试面试订阅2864Exponential Random Intensity Gives Geometric CountsA latent intensity \Lambda is Exponential ( ) with rate . Conditional on \Lambda, the count N is Poisson (\Lambda). Use MGFs to identify the unconditional law of N, and compute E[N].概率中等derivation未尝试面试订阅2865A Stable Law with No MGFSuppose X 1,X 2,\dots are i.i.d. with characteristic function \[ \phi X(u)=\exp(-c|u| 3/2 ),\qquad c>0. \] Show that \[ n -2/3 (X 1+\cdots+X n) \] has the same distribution as X 1.概率困难derivation未尝试面试订阅2866Markov Bound for Daily SlippageA nonnegative slippage random variable L has mean E[L]=2 basis points. Give the best Markov upper bound you can on P(L\ge 10).概率简单derivation未尝试面试订阅2867A Generalized Markov Bound from the Fourth MomentSuppose X is any random variable with E[X 4]=81. Use Markov's inequality on a suitable nonnegative variable to bound P(|X|\ge 6).概率中等derivation未尝试面试订阅2868Recovering Chebyshev from MarkovDerive Chebyshev's inequality from Markov's inequality. In other words, show that for any random variable X with mean and variance 2, \[ P(|X- |\ge a)\le 2 a 2 . \]概率简单derivation未尝试面试订阅2869Chebyshev for a Monte Carlo MeanAn unbiased Monte Carlo estimator averages n=100 i.i.d. draws with variance 9. Use Chebyshev's inequality to bound the probability that the sample mean deviates from its target by at least 0.5.概率简单derivation未尝试面试订阅2870Which Bound Is Better Here?A nonnegative random variable X satisfies E[X]=1 and Var (X)=4. Compare the Markov and Chebyshev upper bounds on P(X\ge 5), and say which one is tighter.概率中等derivation未尝试面试订阅2871Hoeffding for a Bernoulli MeanLet X 1,\dots,X n be i.i.d. Bernoulli(p) and let X n be the sample mean. Use Hoeffding's inequality to bound \[ P( X n-p\ge \varepsilon). \]概率简单derivation未尝试面试订阅2872How Many Bernoulli Trials for 99% Accuracy?You estimate a Bernoulli success probability by the sample mean of n i.i.d. trials. How large must n be to guarantee, via Hoeffding's inequality, that \[ P(| X n-p|\ge 0.02)\le 0.01? \]概率中等derivation未尝试面试订阅2873Hoeffding Versus Chebyshev for 65 Heads in 100 TossesA fair coin is tossed 100 times. Compare the Hoeffding and Chebyshev upper bounds on the event that the fraction of heads is at least 0.65.概率中等derivation未尝试面试订阅2874Hoeffding for Bounded Daily PnLSuppose daily centered PnL increments X 1,\dots,X 50 are independent and each lies almost surely in [-2,3]. Bound \[ P\! ( 1 50 \sum i=1 50 X i\ge 0.5 ) \] using Hoeffding's inequality.概率中等derivation未尝试面试订阅2875Monte Carlo Pricing Error with Bounded PayoffsA Monte Carlo pricer averages 500 i.i.d. discounted payoff samples, each in [0,1]. Use Hoeffding's inequality to bound the probability that the estimated price differs from the true price by at least 0.05.概率简单derivation未尝试面试订阅2876Sub-Gaussian Tail from an MGF AssumptionSuppose a centered random variable X satisfies \[ E[e tX ]\le e 2 t 2/2 \qquad for all t\in R. \] Use exponential Markov to prove that \[ P(X\ge x)\le e -x 2/(2 2) . \]概率中等derivation未尝试面试订阅