Problem 1

Let \(X_1,...,X_{100}\sim Exp(1.3)\), independent, and let \(S=\sum _{i=1}^{100} X_i\). Find \(P(S>100)\)

  1. directly

  2. using the CLT

  3. \(S\sim Gamma(100, 1.3)\) and so

1-pgamma(100, 100, 1.3)
## [1] 0.002750408

\[ \begin{aligned} &E[X_1]=1/1.3\\ &var(X_1)=1/1.3^2\\ &P(S>100) = P(\bar{X}>1)= \\ &P\left(\sqrt{n}\frac{\bar{X}-\mu}{\sigma}>\sqrt{100}\frac{1-1/1.3}{1/1.3}\right)\approx\\ &P(Z>3) = 0.00135 \end{aligned} \]

1-pnorm(3)
## [1] 0.001349898

Problem 2

Let \(X_n\sim U[0,1]\), n=1,2,.., and independent. Use the CLT to find \(P(\sum_{i=1}^{50} X_{i}>24)\). Give an upper and a lower bound for this probability.

\[ \begin{aligned} &EX_n = \frac12 \\ &var X_n = \frac1{12}\\ &P(\sum_{i=1}^{50} X_i > 24) = \\ &P(\bar{X} > 24/50) = \\ &P(\sqrt{50}\frac{\bar{X} - \frac12}{\sqrt{1/12}} > \sqrt{50}\frac{24/50 - \frac12}{\sqrt{1/12}}) \approx \\ &P(Z>-0.49) = 0.688 \end{aligned} \] By the Berry-Essen theorem we have

\[|P(\sum_{i=1}^{50} X_i > 24)-\Phi(-0.49|\le \frac{0.4748\rho}{\sigma^3\sqrt{n}}\]

so

\[ \begin{aligned} &\rho=E|Y_n|^3 =\int_{0}^{1} |x-1/2|^3 dx = \\ &\int_{0}^{1/2} -(x-1/2)^3 dx + \int_{1/2}^{1} (x-1/2)^3 dx = \\ &-(x-1/2)^4/4|_0^{1/2} + (x-1/2)^4/4|_{1/2}^1 = \\ &(0-1/2)^4/4+(1-1/2)^4/4 = \\ &2(1/2)^4/4=(1/2)^5=1/32 \end{aligned} \]

\[ \begin{aligned} &P(\sum_{i=1}^{50} X_i > 24) = \\ &0.688\pm \frac{0.4748/32}{(\sqrt{1/12})^3\sqrt{50}} \\ &0.688\pm 0.087 \end{aligned} \]