Q. 8.7 - A First Course in Probability

Question

Suppose that a fair die is rolled $100$times. Let $X_i$be the value obtained on the $i$th roll. Compute an approximation for$P\left\{\prod _1^{100}{X}_i\le a^{100}\right\} 1 < a < 6$.

Step-by-Step Solution

Answer

$P\left\{\prod _1^{100}{X}_i\le a^{100}\right\}\approx \Phi \left(\frac{100\ln \left(a\right)-100{\mu }_l}{10{\sigma }_l}\right)$

where ${\mu }_l=E\left(\ln \left(X_i\right)\right)$and${\sigma }_l^2=Var\left(\ln \left(X_i\right)\right)$.

Step 1 Given Information.

a fair die is rolled $100$times. $X_i$ be the value obtained on the,$i$th roll.

Step 2 Explanation.

Assume that a fair die is rolled$100$ times and let$X_i$ represent the value obtained on the $i$th roll. Then, the random variables $X_1,X_2,\dots ,X_{100}$are independent identically distributed, with mean

$\mu ={\mu }_i=E\left(X_i\right)=\frac{1}{6}(1+2+3+4+5+6)=\frac{7}{2}$

and variance

${\sigma }^2={\sigma }_i^2=Var\left(X_i\right)=\frac{1}{6}\left(1^2+2^2+3^2+4^2+5^2+6^2\right)-{\mu }^2=\frac{35}{12}$

Therefore, the random variables $\ln \left(X_1\right),\ln \left(X_2\right),\dots ,\ln \left(X_{100}\right)$ are independent identically distributed, with mean

${\mu }_l=E\left(\ln \left(X_i\right)\right)=\frac{1}{6}\left(\ln \right(1)+\ln (2)+\ln (3)+\ln (4)+\ln (5)+\ln (6\left)\right)\approx 1.1$

and variance

${\sigma }_l^2=Var\left(\ln \left(X_i\right)\right)$$=\frac{1}{6}\left(\ln (1{)}^2+\ln (2{)}^2+\ln (3{)}^2+\ln (4{)}^2+\ln (5{)}^2+\ln (6{)}^2\right)$$-{\mu }_l^2\approx 0.37$

Further, let$1<a<6$. We have:

$$\begin{gathered} P\left\{\prod _1^{100}{X}_i\le a^{100}\right\}=P\left\{\ln \left(\prod _1^{100}{X}_i\right)\le \ln \left(a^{100}\right)\right\}= \\ P\left\{\sum _1^{100}\ln \left(X_i\right)\le 100\ln \left(a\right)\right\}= \\ P\left\{\frac{{\sum }_1^{100}\ln \left(X_i\right)-100{\mu }_l}{{\sigma }_l\sqrt{100}}\le \frac{100\ln \left(a\right)-100{\mu }_l}{{\sigma }_l\sqrt{100}}\right\} \end{gathered}$$

The central limit theorem

$\Phi \left(\frac{100\ln \left(a\right)-100{\mu }_l}{10{\sigma }_l}\right)$