某数字货币单日上涨概率为0.55,定义XXX为上涨天数指示变量:X={1上涨0下跌 X = \begin{cases} 1 & \text{上涨} \\ 0 & \text{下跌} \end{cases} X={10上涨下跌
求E(X)E(X)E(X)和Var(X)\text{Var}(X)Var(X)
某期权交易员成功预测方向的概率为0.6。连续进行5次预测:
E(X)=0.55E(X) = 0.55E(X)=0.55
Var(X)=0.55×0.45=0.2475\text{Var}(X) = 0.55 \times 0.45 = 0.2475Var(X)=0.55×0.45=0.2475
(a) (53)0.630.42=0.3456\binom{5}{3}0.6^30.4^2 = 0.3456(35)0.630.42=0.3456
(b) ∑k=45(5k)0.6k0.45−k=0.3370\sum_{k=4}^5 \binom{5}{k}0.6^k0.4^{5-k} = 0.3370∑k=45(k5)0.6k0.45−k=0.3370
(a) (1−0.15)4×0.15=0.0783(1-0.15)^4 \times 0.15 = 0.0783(1−0.15)4×0.15=0.0783
(b) E(X)=1/0.15≈6.67E(X) = 1/0.15 ≈ 6.67E(X)=1/0.15≈6.67次
(a) e−2.5≈0.0821e^{-2.5} ≈ 0.0821e−2.5≈0.0821
(b) λ=7.5\lambda=7.5λ=7.5, 1−∑k=04e−7.57.5kk!≈0.8671 - \sum_{k=0}^4 e^{-7.5}\frac{7.5^k}{k!} ≈ 0.8671−∑k=04e−7.5k!7.5k≈0.867
λ=50×0.012=0.6\lambda=50 \times 0.012=0.6λ=50×0.012=0.6
(a) e−0.60.6≈0.329e^{-0.6}0.6 ≈ 0.329e−0.60.6≈0.329
(b) 1−e−0.6(1+0.6)≈0.1221 - e^{-0.6}(1+0.6) ≈ 0.1221−e−0.6(1+0.6)≈0.122
(a) λ=120/6=20\lambda=120/6=20λ=120/6=20, e−20≈2.06×10−9e^{-20}≈2.06 \times 10^{-9}e−20≈2.06×10−9
(b) E(T100)=100/120×60=50E(T_{100}) = 100/120 \times 60 = 50E(T100)=100/120×60=50秒
(a) E[V]=eμ+σ22Φ(d1)−KΦ(d2)E[V] = e^{μ + \frac{σ^2}{2}}Φ(d_1) - KΦ(d_2)E[V]=eμ+2σ2Φ(d1)−KΦ(d2)
(b) 波动率增加会拉宽分布尾部,增加期权价值
(a) VaR0.95=−(0.001−1.645×0.022)≈3.58%VaR_{0.95} = -(0.001 - 1.645 \times 0.022) ≈ 3.58\%VaR0.95=−(0.001−1.645×0.022)≈3.58%
(b) VaR10D=10×VaR1D≈11.32%VaR_{10D} = \sqrt{10} \times VaR_{1D} ≈ 11.32\%VaR10D=10×VaR1D≈11.32%
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