Question

7. Let X a be random variable with probability density function given by -1 < x < 1 fx(x) otherwise (a) Find the mean u and variance o2 of X (b) Derive the moment generating function of X and state the values for which it is defined (c) For the value(s) at which the moment generating function found in part (b) is (are) not defined, what should the moment generating function be defined as? Justify your answer (d) Let X1, X2, ... be a sequence of independent random variables where each one has the same distribution as X. Derive the moment generating function of X1X2 + Xn Sn .. (e) Use the moment generating function of Sn to prove that Sn converges to a normal distribution in distribution as n -> 00, and specify the parameters of the limiting distribution. (f) Let X2 +...+ X100 - T X1 Give an approximate value of P(T < 13).

Answer 1

(a)

The mean and variance are given by:

$\begin{align*} \mu &= E(X) \\ &= \int_{-1}^{1} x * f_X(x) \ dx \\ &= \int_{-1}^{1} x * \frac{1}{2} \ dx \\ &= \begin{bmatrix} \frac{x^2}{4}\end{bmatrix}_{x=-1}^{x=1} \\ &= \frac{1^2}{4} - \frac{(-1)^2}{4} \\ &= \frac{1}{4} - \frac{1}{4} \\ &= \textbf{0 \ \ \ \ \ [Answer]} \end{align*}$

$\begin{align*} E(X^2) &= \int_{-1}^{1} x^2 * f_X(x) \ dx \\ &= \int_{-1}^{1} x^2 * \frac{1}{2} \ dx \\ &= \begin{bmatrix} \frac{x^3}{6}\end{bmatrix}_{x=-1}^{x=1} \\ &= \frac{1^3}{6} - \frac{(-1)^3}{6} \\ &= \frac{1}{6} + \frac{1}{6} \\ &= \frac{1}{3} \\ \sigma^2 &= E(X^2) - [E(X)]^2 \\ &= \frac{1}{3} - 0^2 \\ &= \mathbf{\frac{1}{3} \ \ \ \ \ [Answer]} \end{align*}$

(b)

The moment generating function of X is given y:

$\begin{align*} M_X(t) &= \int_{-1}^{1} e^{tx} f_X(x) \ dx \\ &= \int_{-1}^{1} e^{tx} *\frac{1}{2} \ dx \\ &= \frac{1}{2}\int_{-1}^{1} e^{tx} \ dx \\ &= \frac{1}{2}\begin{bmatrix} \frac{e^{tx}}{t} \end{bmatrix}_{x=-1}^{x=1} \\ &= \mathbf{\frac{e^{t} - e^{-t}}{2t}\ \ ; t \ne 0 \ \ \ \ [Answer]} \end{align*}$

The MGF is defined for all real values of t except zero because if we put the value of t=0 in the above formula, we get 0/0 indeterminate form.

(c)

From part (b) we see that the MGF is not defined when t=0. Now, to find what the value of the MGF at t=0, we can use a different approach. We put t=0 first and then integrate to find M_X(0):

$\begin{align*} M_X(0) &= \int_{-1}^{1} e^{0*x} f_X(x) \ dx \\ &= \int_{-1}^{1} 1* f_X(x) \ dx \\ &= \int_{-1}^{1} f_X(x) \ dx \\ &= \textbf{1 \ \ \ \ \ \ [Answer]} \end{align*}$

(d)

$\begin{align*} M_{S_n} (t) &= E[e^{tS_n}] \\ &= E[e^{t(\frac{X_1+...+X_n}{\sqrt{n}})}] \\ &= E[e^{\frac{tX_1}{\sqrt{n}}} * e^{\frac{tX_2}{\sqrt{n}}} * ... * e^{\frac{tX_n}{\sqrt{n}}}] \\ &= E[e^{\frac{tX_1}{\sqrt{n}}}] * E[e^{\frac{tX_2}{\sqrt{n}}} ]* ... * E[e^{\frac{tX_n}{\sqrt{n}}}] \ \ \ \ \text{(Since, X}_i's \text{ are independent)} \\ &= M_{X_1}(t/\sqrt{n})*M_{X_2}(t/\sqrt{n})*...*M_{X_n}(t/\sqrt{n}) \\ &= [M_{X_1}(t/\sqrt{n})]^n \ \ \ \ \text{(Since, X}_i's \text{ are identically distributed)} \\ &= \mathbf{\begin{bmatrix} \frac{e^{t/\sqrt{n}}-e^{-t/\sqrt{n}}}{2(t/\sqrt{n})} \end{bmatrix}^n \ \ \ \ [Answer]} \end{align*}$

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