Q1: The planet Tralfamadore has years with 500 days. There are 5 Tralfamadorans in the room. Write an expression for the probability that no two of them have the same birthday.
Solution: Suppose that there are n Tralfamadorans in the room. The probability that no two of them have the same birthday is given by
p(n):={
A500n500n,n≤5000,n>500 p(n):= \begin{cases} \frac{A_{500}^n}{500^n}, & n\leq500\\ 0,&n>500 \end{cases} p(n):={
500nA500n,0,n≤500n>500
When n=5n=5n=5, we have p(5)=A5005/5005≈0.9801p(5)=A_{500}^5/500^5\approx0.9801p(5)=A5005/5005≈0.9801
How would you find the smallest nnn for which a room of nnn Tralfamadorans has probability at least 1/21/21/2 of having two members with the same birthday?
Solution: nnn要满足的条件为
1−500!(500−n)!500n≥12 1-\frac{500!}{(500-n)!500^n}\geq\frac{1}{2} 1−(500−n)!500n500!≥21
即
500!(500−n)!500n≤12 \frac{500!}{(500-n)!500^n}\leq\frac{1}{2} (500−n)!500n500!≤21
可计算得nnn的最小值为27
The above two questions really require some sort of assumption to get an answer. In case you did not already provide one, what is the customary assumption one uses in probability exercises?
Solution: The birthdays of the nnn Tralfamadorans are independent and uniformly distributed over the 500 days.
Q2: Write an expression for ϕX(t)\phi_X(t)ϕX(t), the moment generating function (MGF) of a random variable X. Find and interpret the second derivative ϕ′′X(0)\phi{''}_X(0)ϕ′′X(0). If the MGF does not exist what would we use instead?
Solution:
ϕx(t)=∫−∞+∞etxdF(x) \phi_x(t)=\displaystyle\int_{-\infty}^{+\infty}e^{tx}dF(x) ϕx(t)=∫−∞+∞etxdF(x)
ϕx′′(0)=∫−∞+∞x2etxdF(x)∣t=0=Ex2 \phi_x^{''}(0)=\displaystyle\int_{-\infty}^{+\infty}x^2e^{tx}dF(x)|_{t=0}=Ex^2 ϕx′′(0)=∫−∞+∞x2etxdF(x)∣t=0=Ex2
用特征函数表示
ϕx′′(0)=Ex2=−f2(0) \phi_x^{''}(0)=Ex^2=-f^{2}(0) ϕx′′(0)=Ex2=−f2(0)
Q3: If XXX and YYY are uncorrelated random variables must they be independent?If XXX and YYY are independent random variables must they be uncorrelated?Explain in both cases.
Solution: 如果XXX和YYY是不相关的随机变量,它们不一定是独立的,比如说,存在一个简单的例子,
X \ Y | -1 | 0 | 1 |
---|---|---|---|
0 | 0 | 1/3 | 0 |
1 | 1/3 | 0 | 1/3 |
从表格中可以得知,Eξ=0E\xi=0Eξ=0, Eη=Eξ2=23E\eta=E\xi^2=\frac{2}{3}Eη=Eξ2=32, Eξη=Eξ3=0E\xi\eta=E\xi^3=0Eξη=Eξ3=0,故cov(ξ,η)=Eξη−Eξ⋅Eη=0cov(\xi,\eta)=E\xi\eta-E\xi{\cdot}E\eta=0cov(ξ,η)=Eξη−Eξ⋅Eη=0,所以ξ\xiξ与η\eta