Stat 5131 Midterm 2

N = 24   Median = 68.5
Quartiles = 47, 87          grad       undergrad

    3 : 14                             3 : 14           
    3 : 5                              3 : 5            
    4 : 2                              4 : 2            
    4 : 577                            4 : 577          
    5 : 4                              5 : 4            
    5 :                                5 :              
    6 : 223                 6 : 3      6 : 22
    6 :                     6 :        6 :
    7 : 1144                7 : 1      7 : 144
    7 : 56                  7 : 5      7 : 6 
    8 : 2                   8 :        8 : 2 
    8 :                     8 :        8 :   
    9 : 24                  9 : 4      9 : 2 
    9 : 57                  9 : 57     9 :   
   10 : 00                 10 : 0     10 : 0

Problem 1

Median = 20, Lower Quartile = 19, Upper Quartile = 20

(a)

The distribution of X is symmetric about zero, hence E(X) = 0by Theorem 4 of Chapter 4 in Lindgren.

(b)

Since E(X) = 0 by part (a),
$$\begin{align*}\mathop{\rm var}\nolimits(X) & = E(X^2) \\ & = \int x^2 f(x)... ...4} \left( \frac{1}{3} - \frac{1}{7} \right) \\ & = \frac{5}{21} \end{align*}$$

Problem 2

Median = 20, Lower Quartile = 16, Upper Quartile = 20

The conditional p. d. f. of X given Y is

$$f(x \vert y) = \frac{f(x, y)}{f_Y(y)}$$

So to find f(x | y) we first have to find the marginal f_Y(y). (Note: this was done in problem 4 of the first midterm. This part of the question is a rehash.)

To find the marginal of Y, integrate out X
$$\begin{align*}f_Y(y) & = \int_0^1 \tfrac{6}{7} (x + y)^2 \, d x \\ & = \tf... ...t _0^1 \\ & = \tfrac{6}{7} \left(y^2 + y + \tfrac{1}{3} \right) \end{align*}$$
Then
$$\begin{align*}f(x \vert y) & = \frac{f(x, y)}{f_Y(y)} \\ & = \frac{\tfrac{... ... \tfrac{1}{3})} \\ & = \frac{(x + y)^2}{y^2 + y + \tfrac{1}{3}} \end{align*}$$

Problem 3

Median = 0, Lower Quartile = 0, Upper Quartile = 16.5

The regression function of Y on X is another name for E(Y | X).
$$\begin{align*}E(Y \vert X = x) & = \int y f(y \vert x) \, d y \\ & = \int_... ...2 e^{- y} \, d y + \frac{x}{1 + x} \int_0^\infty y e^{- y} \, d y \end{align*}$$
From the formula given in the hint

$$\int_0^\infty y^2 e^{- y} \, d y = 2 \qquad \text{and} \qquad \int_0^\infty y e^{- y} \, d y = 1$$

so the regression function is

$$E(Y \vert X = x) = \frac{2 + x}{1 + x}, \qquad x > 0.$$

Problem 4

Median = 19, Lower Quartile = 9, Upper Quartile = 20

(a)

$$\begin{align*}\mathop{\rm var}\nolimits(X + Y) & = \mathop{\rm var}\nolimits(X... ...p{\rm cov}\nolimits(X, Y) \\ & = 1 + 1 + 2 (- 0.5) \\ & = 1 \end{align*}$$

(b)

$$\begin{align*}\mathop{\rm var}\nolimits(X + Y + Z) & = \mathop{\rm var}\nolimi... ...its(Z, X) \\ & = 1 + 1 + 1 + 2 (- 0.5 - 0.5 - 0.5) \\ & = 0 \end{align*}$$

Problem 5

Median = 11.5, Lower Quartile = 4, Upper Quartile = 16

$$\begin{align*}\psi'(t) & = \frac{(1 - p) p e^t}{(1 - p e^t)^2} \\ \psi''(t) & ... ... p e^t}{(1 - p e^t)^2} + \frac{2 (1 - p) (p e^t)^2}{(1 - p e^t)^3} \end{align*}$$
Evaluating at zero we get
$$\begin{align*}E(X) & = \psi'(0) = \frac{(1 - p) p}{(1 - p)^2} = \frac{p}{1... ... - p) p^2}{(1 - p)^3} = \frac{p}{1 - p} + \frac{2 p^2}{(1 - p)^2} \end{align*}$$
If we write $E(X) = \mu$, then

$$E(X^2) = \mu + 2 \mu^2$$

and

$$\mathop{\rm var}\nolimits(X) = E(X^2) - \mu^2 = \mu + \mu^2 ... ...frac{p}{1 - p} + \frac{p^2}{(1 - p)^2} = \frac{p}{(1 - p)^2}$$