The price of one share of stock in the Pilsdorff Beer Company (see Exercise 8.2.12) is given by \(Y_n\) on the nth day of the year. Finn observes that the differences \(X_n = Y_{n+1} − Y_n\) appear to be independent random variables with a common distribution having mean \(\mu = 0\) and variance \(\sigma_2 = 1/4\). If \(Y_1 = 100\), estimate the probability that \(Y_{365}\) is
To apply the Central Limit Theorem, we can’t use \(Y_n\) but need to express the probability for \(Y_n\) in terms of a sum \(S_n\) of i.i.d random variables. Using simple algebra, we rearrange \(Y_n\) as a telescoping series and use the identity \(X_i = Y_{i+1}-Y_{i}\).
\[\begin{align} Y_n &= & Y_1 + (Y_2 - Y_1)+(Y_3-Y_2) + \dots + (Y_n - Y_{n-1}) \\ & = & Y_1 + X_1 + X_2 + \dots + X_{n-1} \\ Y_n - Y_1 & = & X_1 + X_2 + \dots + X_{n-1}\\ Y_n - Y_1 & = & S_{n-1} \\ \end{align} \]
It follows that for \(a < b\), by the CLT,
\[ lim_{n \rightarrow \infty } P\left[ a < \frac{ S_n - n \mu}{\sqrt{n}\sigma} < b \right] = \Phi(b) - \Phi(a)\]
We use the same approach to solve (a)-(c) by formulating a lemma stated below:
For any real \(C\), \[ P\left[ Y_{365} > 100 + C \right] \sim \Phi( -\frac{2C}{\sqrt{364}}) \] Proof of Lemma
We know that \(Y_1 = 100\) and
\[\begin{align} P[Y_n > 100 + C ] & = & P[Y_n > Y_1 + C] \\ & = & P[ Y_n - Y_1 > C ] \\ & = & P[ S_{n-1} > C ] \\ \end{align} \] By the CLT, the latter term is equivalent by minor algebra re-arrangement to:
\[\begin{align} P[S_{n-1} > C ] = P\left[ \frac{ S_{n-1} - (n-1)\mu}{ \sqrt{n-1}\sigma} > \frac{ C - (n-1)\mu}{\sqrt{n-1}\sigma} \right] \sim 1 - \Phi(\frac{C - (n-1)\mu}{\sqrt{n-1}\sigma}) \end{align} \]
Since \(\mu = 0\) and \(n=365\) and \(\sigma = 1/2\), we obtain the equivalent format: \[P\left[ Y_{365} > 100 + C \right] \sim \Phi( -\frac{C}{\sqrt{364} \frac{1}{2} } ) = \Phi(- \frac{ 2C}{\sqrt{364}}) \]
This proves the lemma.
To solve (a), set \(C=0\). To solve (b), set \(C=10\) and to solve (c), set \(C = 20\).
We calculate the estimates in R:
(prob_100 = pnorm( -2 * 0 / sqrt(364) ) )## [1] 0.5(prob_110 = pnorm( -2 * 10/ sqrt(364) ) )## [1] 0.1472537(prob_120 = pnorm( -2 * 20/ sqrt(364) ) )## [1] 0.01801584We estimate \[P[ Y_{365} > 100 ] \approx 50.000\%\]
We estimate \[P[ Y_{365} > 110 ] \approx 14.725\%\]
We estimate \[P[ Y_{365} > 120 ] \approx 1.802\%\]
Calculate the expected value and variance of the binomial distribution using the moment generating function.
The moment generating function of the binomial distribution with parameters \(n\) and \(p\) is:
\[m(t) = (1 - p + pe^t)^n\] Using the identity \[E[X^k] = \frac{\partial^{(k)} m}{\partial t}(0)\]
We solve for the first and second derivatives with respect to \(t\):
\[\begin{align} m^{'}(t) & = & n ( 1 - p + pe^t)^{n-1}(pe^t) \\ m^{''}(t) & = & n (n-1) ( 1 - p + pe^t)^{n-2}(pe^t)^2 + n(1-p + pe^t)^{n-1}pe^t \\ \end{align} \]
Noting that when \(t = 0\), we get $ (1 - p + pe^0) = 1$ and \(pe^0 = p\), the above derivatives can be evaluated to:
\[ \begin{align} m^{'}(0) & = & E[X^1] & = & np \\ m^{''}(0) & = & E[X^2] & = & n(n-1)p^2 + np \\ \end{align} \]
The mean is of the binomial distribution is therefore \[E[X] = np\]
The variance is obtained from the identity: \[\begin{align} E[X^2] - (E[X]^2)& = & n(n-1)p^2 + np - n^2p^2 \\ V(X) & = & n^2p^2 -np^2 + np - n^2p^2 \\ & = & np - np^2 \\ & = & np(1-p) \\ \end{align} \]
Thus, the variance of the binomial distribution is \(V(X) = np(1-p)\).
Calculate the expected value and variance of the exponential distribution using the moment generating function.
The moment generating function of the exponential distribution is \(m(t)\) where
\[ m(t) = \frac{ \lambda}{\lambda - t} \text{ for } t < \lambda \]
Calculating the first derivative of \(m(t)\) with respect to \(t\), we obtain:
\[\begin{align} \frac{\partial m}{\partial t} & = & \lambda (-1)(\lambda - t)^{-2} (-1) \\ & = & \lambda ( \lambda - t)^{-2} \\ \end{align} \]
Setting \(t=0\), we get
\[\frac{\partial m}{\partial t}(0) = E[X] = \lambda ( \lambda - 0)^{-2} = \frac{1}{\lambda}\]
Thus the mean of the exponential distribution is \[E[X] = \frac{1}{\lambda}\]
Taking the second derivative of \(m(t)\) with respect to \(t\) we get:
\[\begin{align} \frac{\partial^2 m}{\partial t} & = & \lambda (-2)(\lambda - t)^{-3}(-1) \\ & = & \frac{ 2\lambda}{(\lambda - t)^3} \\ \end{align} \]
Plugging in \(t = 0\) gives
\[\frac{\partial^2 m}{\partial t} (0) = E[X^2] = \frac{2}{\lambda^2}\]
Lastly, using the identity for the variance \(V(X)\):
\[\begin{align} V(X) & = E[X^2] - E[X]^2 \\ & = \frac{2}{\lambda^2 } - (\frac{1}{\lambda})^2 \\ & = \frac{1}{\lambda^2} \end{align} \]
Thus, the variance of the exponential distribution is \[V(X) = \frac{1}{\lambda^2}\].