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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2877Rademacher-Sum Upper TailLet X 1,\dots,X 100 be i.i.d. with P(X i=1)=P(X i=-1)=1/2. Use a Chernoff-style bound to estimate \[ P (\sum i=1 100 X i\ge 20 ). \]概率中等derivation未尝试面试订阅2878Poisson Upper Tail in Multiplicative FormLet N\sim Poisson ( ). Show that for any >0, \[ P(N\ge (1+ ) )\le \exp\! (- \bigl((1+ )\ln(1+ )- \bigr) ). \]概率中等derivation未尝试面试订阅2879Numerical Poisson Overload BoundAn exchange gateway receives N\sim Poisson (100) messages in a fixed interval. Use the Poisson Chernoff upper-tail bound to estimate P(N\ge 130).概率中等derivation未尝试面试订阅2880Poisson Lower Tail BoundLet N\sim Poisson ( ). Show that for 0< <1, \[ P(N\le (1- ) )\le \exp\! (- \bigl( +(1- )\ln(1- )\bigr) ). \]概率中等derivation未尝试面试订阅2881A Numerical Poisson Shortfall BoundIf N\sim Poisson (100), use the lower-tail Chernoff bound to estimate P(N\le 80).概率中等derivation未尝试面试订阅2882Optimizing a Chernoff Bound for an Exponential VariableLet X\sim Exponential (1). Use its MGF to derive the best Chernoff-type upper bound you can on P(X\ge a) for a>1.概率中等derivation未尝试面试订阅2883A Chernoff Bound for a Sum of ExponentialsLet S=X 1+\cdots+X k where X i\overset i.i.d. \sim Exponential (1). Use the MGF to derive a Chernoff upper bound for P(S\ge a) when a>k.概率困难derivation未尝试面试订阅2884Multiplicative Chernoff for a Binomial CountLet X\sim Binomial (n,p) with mean =np. Show that for any >0, \[ P(X\ge (1+ ) )\le ( e (1+ ) 1+ ) . \]概率困难derivation未尝试面试订阅2885Sub-Gaussian Sum with a Volatility ProxySuppose X 1,\dots,X n are independent centered random variables and each satisfies \[ E[e tX i ]\le e 2 t 2/2 \qquad for all t\in R. \] Show that for S n=\sum i=1 n X i, \[ P(S n\ge x)\le \exp\! (- x 2 2n 2 ). \]概率中等derivation未尝试面试订阅