\(P(Y_{365} - Y_1 \geq 0)\)
a <- 0/sqrt(364)
sigma <- sqrt(1/4)
1 - pnorm(q = a, mean = 0, sd = sigma, lower.tail = TRUE)
## [1] 0.5
a <- 10/sqrt(364)
sigma <- sqrt(1/4)
1 - pnorm(q = a, mean = 0, sd = sigma, lower.tail = TRUE)
## [1] 0.1472537
a <- 20/sqrt(364)
sigma <- sqrt(1/4)
1 - pnorm(q = a, mean = 0, sd = sigma, lower.tail = TRUE)
## [1] 0.01801584
2.Calculate the expected value and variance of the binomial distribution using the moment generating function The binomial distrubution \(P(X=k) = {n \choose k} p^k q^{n-k}\) when \(q=1-p\)
The moment generating function is \(M_X(t)=(q+pe^t)^n\)
The first moment function is \(M′X(t)=n(q+pet)n−1pet\)
\[\begin{split} E(X)=M'_X(0) &= n(q+pe^0)^{n-1}pe^0\\ &= n(q+p)^{n-1}p\\ &= np(1-p+p)^{n-1}\\ &= np1^{n-1}\\ &=np \end{split}\]
The second moment function is \(M''_X(t) = n(n-1)(q+pe^t)^{n-2}p^2 e^{2t}+n(q+pe^t)^{n-1}pe^t\)
\(E(X2)=M″X(0)=n(n−1)(q+pe0)n−2p2e0+n(q+pe0)n−1pe0=n(n−1)(1−p+p)n−2p2+n(1−p+p)n−1p=n(n−1)p2+np\)
The moment generating function is: \(MX(t)=λλ−t,t<λ\)
The expected value is: \[\begin{split} E(X)=M'_X(0) &= \frac{\lambda}{(\lambda-0)^2} \\ &= \frac{\lambda}{\lambda^2}\\ &= \frac{1}{\lambda} \end{split}\]
The variance is: \[\begin{split} V(X) = E(X^2)-E(X)^2 &= M''_X(0)-M'_X(0)^2 \\ &=\frac{2\lambda}{(\lambda-0)^3} - \frac{1}{\lambda^2}\\ &=\frac{2\lambda}{\lambda^3} - \frac{1}{\lambda^2}\\ &=\frac{2}{\lambda^2} - \frac{1}{\lambda^2}\\ &=\frac{1}{\lambda^2} \end{split}\]