First, let’s define our distribution parameters for \(X_n = Y_{n + 1} - Y_{n}\). We can do some manipulation and substitution to get to a random variable definition we can use:
\[\begin{aligned} X_n = Y_{n + 1} - Y_{n} Y_{n+1} = X_n + Y_n \end{aligned}\]mu <- 0
sigma = sqrt(1 / 4)
sn <- sqrt(365)(pnorm(q = (100 - 100)/sn, mean = mu, sd = sigma, lower.tail = TRUE))## [1] 0.5(pnorm(q = (100 - 110)/sn, mean = mu, sd = sigma, lower.tail = TRUE))## [1] 0.1475849(pnorm(q = (100 - 120)/sn, mean = mu, sd = sigma, lower.tail = TRUE))## [1] 0.01814355We start with a binomial distribution for\(k\) successes in \(n\) trials with probability of success \(p\)
\[\begin{aligned} P(k: n, p) = {n \choose k}p^kq^{n-k} \end{aligned}\]where \(q = 1 - p\). Since these trials are discrete, we can use the moment-generating function for discrete probability distributions:
\[\begin{aligned} M_x(t) = \sum_{x}e^{tx}P(x) \end{aligned}\]Plugging in our binomial probability distribution, we get:
\[\begin{aligned} M_k(t) = \sum_{k}e^{tk}{n \choose k}p^kq^{n-k} \newline = \sum_{k}{n \choose k}(pe^t)^kq^{n-k} \end{aligned}\]Here we make use of the binomial theorem, which states:
\[\begin{aligned} (x + y)^n = \sum_{k=0}^{n}{n \choose k}x^ky^{n-k} \end{aligned}\]Inserting this into our formula, we get:
\[\begin{aligned} M_k(t) = \sum_{k}{n \choose k}(pe^t)^kq^{n-k} \newline M_k(t) = (pe^t + q)^n \end{aligned}\]Now that we have our moment generator function \(M_k(t)\), we can differentiate to get our expected value and variance of our binomial distribution:
\[\begin{aligned} E(X) = \mu_1 = M_k'(0) = n(pe^t + q)^{n-1}pe^t \rvert_{t=0} \newline = n(pe^0 + 1 - p)^{n-1}pe^0 \newline = np \end{aligned}\]This result aligns with what we know about the properties of the binomial distribution. Now to find the variance, we’ll need to differentiate again. We also need to remember the definition of our vairance based off moments:
\[\begin{aligned} \sigma^2 = E(X^2) - E(X)^2 = \mu_2 - \mu_{1}^{2} \end{aligned}\]For the second moment, we calculate:
\[\begin{aligned} Var(X) = \mu_2 = M_k''(0) = \frac{d}{\,dt}(n(pe^t + q)^{n-1}pe^t) \rvert_{t=0} \end{aligned}\]We can use the product rule to find the second derivative:
\[\begin{aligned} M_k''(0) = n(n-1)(pe^t + q)^{n - 2}p^2e^2t + n(pe^t + q)^{n-1}pe^t \rvert_{t=0} \newline = n(n-1)(pe^0 + q)^{n - 2}p^2e^0 + n(pe^t + q)^{n-1}pe^0 \newline = n(n-1)p + np \newline = n^2p^2 - np^2 + np \end{aligned}\]Now, we can plug the above values into our variance formula \(Var(x) = \sigma^2 = \mu_2 - \mu_1^2\)
\[\begin{aligned} \sigma^2 = n^2p^2 - np^2 + np - n^2p^2 \newline = np - np^2 = np(1 - p) \end{aligned}\]Which again aligns with the properties we know of the binomial distribution
We know based on the definition of a random variable \(X\) having an exponential distribution that the density function looks like:
\[\begin{aligned} f_X(x) = \begin{cases} \lambda e^{-\lambda x} & \text{if } x \ge 0\\ 0 &otherwise \end{cases} \end{aligned}\]We first calculate the moment generating function:
\[\begin{aligned} M_x(t) = \int_{0}^{\infty}e^{tx}f(x)\,dx \newline = \int_{0}^{\infty}e^{tx} \lambda e^{-\lambda x}\,dx \newline = \lambda \int_{0}^{\infty}e^{tx} e^{-\lambda x}\,dx \newline = \lambda \int_{0}^{\infty} e^{x(t -\lambda )}\,dx \end{aligned}\]Integrating with respect to \(x\), we get
\[\begin{aligned} M_x(t) = \frac{-\lambda}{t - \lambda} e^{x(t - \lambda)} \rvert_{0}^{\infty} \newline = \frac{-\lambda}{t - \lambda}\left ( e^{\infty(t - \lambda)} - e^0\right) \end{aligned}\]The above expression converges only in the case where \(t < \lambda\), because then the exponential argument in the first term does not diverge. Taking this constraint into account, we can simplify to get:
\[\begin{aligned} M_x(t) = frac{-\lambda}{t - \lambda}, t < \lambda \end{aligned}\]We can differentiate to find the first moment (and consequently the expected value of our exponential distribution):
\[\begin{aligned} E(X) = \mu_1 = M_x'(0) = -\lambda \frac{d}{\,dt}((t - \lambda)^-1) \rvert_{t = 0} \newline = \frac{\lambda}{(t - \lambda)^2} \rvert_{t = 0} = \frac{\lambda}{\lambda^2} = \frac{1}{\lambda} \end{aligned}\]Which aligned with what we know about exponential distributions
We now can find the variance by finding \(\mu_2\), then plugging into our variance formula: \(\sigma^2 = \mu_2 - \mu_1^2\)
\[\begin{aligned} \mu_2 = M_x''(0) = \frac{d}{\,dt}\left[ \frac{\lambda}{(t - \lambda)^2}\right] \rvert_{t=0} \newline = \frac{-2\lambda}{(t - \lambda)^3} \rvert_{t=0} \newline = \frac{-2\lambda}{\lambda^3} \newline = \frac{2}{\lambda^2} \end{aligned}\]Now we can get our variance value
\[\begin{aligned} Var(x) = \sigma^2 = \mu_2 - \mu_1^2 = \frac{2}{\lambda^2} - (\frac{1}{\lambda})^2 = \frac{1}{\lambda^2} \end{aligned}\]