14.1 Problems on conditional expectation, regression (Page 3/5)

Applied probability Page 3 / 5

(See Exercise 18 from " Problems On Random Vectors and Joint Distributions", Exercise 28 from "Problems on Mathematical Expectation", and Exercise 28 from "Problems on Variance, Covariance, Linear Regression"). $f_{X Y} (t, u) = \frac{3}{23} (t + 2 u)$ for $0 \leq t \leq 2$ , $0 \leq u \leq max {2 - t, t}$ .

The regression line of Y on X is $u = 1.0561 t - 0.2603$ .

f_{X} (t) = I_{[0, 1]} (t) \frac{6}{23} (2 - t) + I_{(1, 2]} (t) \frac{6}{23} t^{2}

The regression line of Y on X is $u = 1.0561 t - 0.2603$ .

f_{Y | X} (u | t) = I_{[0, 1]} (t) \frac{t + 2 u}{2 (2 - t)} + I_{(1, 2]} (t) \frac{t + 2 u}{2 t^{2}} 0 \leq u \leq max (2 - t, t)

E [Y | X = t] = I_{[0, 1]} (t) \frac{1}{2 (2 - t)} \int_{0}^{2 - t} (t u + 2 u^{2}) d u + I_{(1, 2]} (t) \frac{1}{2 t^{2}} \int_{0}^{t} (t u + 2 u^{2}) d u

= I_{[0, 1]} (t) \frac{1}{12} (t - 2) (t - 8) + I_{(1, 2]} (t) \frac{7}{12} t

tuappr:  [0 2] [0 2]200 200 (3/23)*(t+2*u).*(u<=max(2-t,t))
EYx = sum(u.*P)./sum(P);M = X<=1;
eyx = (1/12)*(X-2).*(X-8).*M + (7/12)*X.*(1-M);plot(X,EYx,X,eyx)             % Plots quite close

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(See Exercise 21 from " Problems On Random Vectors and Joint Distributions", Exercise 31 from "Problems on Mathematical Expectation", and Exercise 29 from "Problems on Variance, Covariance, Linear Regression"). $f_{X Y} (t, u) = \frac{2}{13} (t + 2 u)$ , for $0 \leq t \leq 2$ , $0 \leq u \leq min {2 t, 3 - t}$ .

The regression line of Y on X is $u = - 0.1359 t + 1.0839$ .

f_{X} (t) = I_{[0, 1]} (t) \frac{12}{13} t^{2} + I_{(1, 2]} (t) \frac{6}{13} (3 - t)

The regression line of Y on X is $u = - 0.1359 t + 1.0839$ .

f_{Y | X} (t | u) = I_{[0, 1]} (t) \frac{t + 2 u}{6 t^{2}} + I_{(1, 2]} (t) \frac{t + 2 u}{3 (3 - t)} 0 \leq u \leq max (2 t, 3 - t)

E [Y | X = t] = I_{[0, 1]} (t) \frac{1}{6 t^{2}} \int_{0}^{t} (t u + 2 u^{2}) d u + I_{(1, 2]} (t) \frac{1}{3 (3 - t)} \int_{0}^{3 - t} (t u + 2 u^{2}) d u

= I_{[0, 1]} (t) \frac{11}{9} t + I_{(1, 2]} (t) \frac{1}{18} (t^{2} - 15 t + 36)

tuappr: [0 2] [0 2]200 200 (2/13)*(t+2*u).*(u<=min(2*t,3-t))
EYx = sum(u.*P)./sum(P);M = X<=1;
eyx = (11/9)*X.*M + (1/18)*(X.^2 - 15*X + 36).*(1-M);plot(X,EYx,X,eyx)              % Plots quite close

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(See Exercise 22 from " Problems On Random Vectors and Joint Distributions", Exercise 32 from "Problems on Mathematical Expectation", and Exercise 30 from "Problems on Variance, Covariance, Linear Regression"). $f_{X Y} (t, u) = I_{[0, 1]} (t) \frac{3}{8} (t^{2} + 2 u) + I_{(1, 2]} (t) \frac{9}{14} t^{2} u^{2}$ ,

for $0 \leq u \leq 1$ .

The regression line of Y on X is $u = 0.0817 t + 0.5989$ .

f_{X} (t) = I_{[0, 1]} (t) \frac{3}{8} (t^{2} + 1) + I_{(1, 2]} (t) \frac{3}{14} t^{2}

The regression line of Y on X is $u = 0.0817 t + 0.5989$ .

f_{Y | X} (t | u) = I_{[0, 1]} (t) \frac{t^{2} + 2 u}{t^{2} + 1} + I_{(1, 2]} (t) 3 u^{2} 0 \leq u \leq 1

E [Y | X = t] = I_{[0, 1]} (t) \frac{1}{t^{2} + 1} \int_{0}^{1} (t^{2} u + 2 u^{2}) d u + I_{(1, 2]} (t) \int_{0}^{1} 3 u^{3} d u

= I_{[0, 1]} (t) \frac{3 t^{2} + 4}{6 (t^{2} + 1)} + I_{(1, 2]} (t) \frac{3}{4}

tuappr: [0 2] [0 1]200 100 (3/8)*(t.^2 + 2*u).*(t<=1) + ...
       (9/14)*t.^2.*u.^2.*(t>1)
EYx = sum(u.*P)./sum(P);M = X<=1;
eyx = M.*(3*X.^2 + 4)./(6*(X.^2 + 1)) + (3/4)*(1 - M);plot(X,EYx,X,eyx)              % Plots quite close

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For the distributions in Exercises 12-16 below

Determine analytically $E [Z | X = t]$
Use a discrete approximation to calculate the same functions.

$f_{X Y} (t, u) = \frac{3}{88} (2 t + 3 u^{2})$ for $0 \leq t \leq 2$ , $0 \leq u \leq 1 + t$ (see Exercise 37 from "Problems on Mathematical Expectation", and [link] ).

f_{X} (t) = \frac{3}{88} (1 + t) (1 + 4 t + t^{2}) = \frac{3}{88} (1 + 5 t + 5 t^{2} + t^{3}), 0 \leq t \leq 2

Z = I_{[0, 1]} (X) 4 X + I_{(1, 2]} (X) (X + Y)

$Z = I_{M} (X) 4 X + I_{N} (X) (X + Y)$ , Use of linearity, (CE8) , and (CE10) gives

E [Z | X = t] = I_{M} (t) 4 t + I_{N} (t) (t + E [Y | X = t])

= I_{M} (t) 4 t + I_{N} (t) (t + \frac{(t + 1) (t + 3) (3 t + 1)}{4 (1 + 4 t + t^{2})})

% Continuation of
[link] G = 4*t.*(t<=1) + (t + u).*(t>1);
EZx = sum(G.*P)./sum(P);M = X<=1;
ezx = 4*X.*M + (X + (X+1).*(X+3).*(3*X+1)./(4*(1 + 4*X + X.^2))).*(1-M);plot(X,EZx,X,ezx)              % Plots nearly indistinguishable

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$f_{X Y} (t, u) = \frac{24}{11} t u$ for $0 \leq t \leq 2$ , $0 \leq u \leq min {1, 2 - t}$ (see Exercise 38 from "Problems on Mathematical Expectaton", [link] ).

f_{X} (t) = I_{[0, 1]} (t) \frac{12}{11} t + I_{(1, 2]} (t) \frac{12}{11} t {(2 - t)}^{2}

Z = I_{M} (X, Y) \frac{1}{2} X + I_{M^{c}} (X, Y) Y^{2}, M = {(t, u) : u > t}

Z = I_{M} (X, Y) \frac{1}{2} X + I_{M^{c}} (X, Y) Y^{2}, M = {(t, u) : u > t}

I_{M} (t, u) = I_{[0, 1]} (t) I_{[t, 1]} (u) I_{M^{c}} (t, u) = I_{[0, 1]} (t) I_{[0, t]} (u) + I_{(1, 2]} (t) I_{[0, 2 - t]} (u)

E [Z | X = t] = I_{[0, 1]} (t) [\frac{t}{2} \int_{t}^{1} 2 u d u + \int_{0}^{t} u^{2} \cdot 2 u d u] + I_{(1, 2]} (t) \int_{0}^{2 - t} u^{2} \cdot \frac{2 u}{{(2 - t)}^{2}} d u

= I_{[0, 1]} (t) \frac{1}{2} t (1 - t^{2} + t^{3}) + I_{(1, 2]} (t) \frac{1}{2} {(2 - t)}^{2}

% Continuation of
[link] Q = u>t;
G = (1/2)*t.*Q + u.^2.*(1-Q);EZx = sum(G.*P)./sum(P);
M = X<= 1;
ezx = (1/2)*X.*(1-X.^2+X.^3).*M + (1/2)*(2-X).^2.*(1-M);plot(X,EZx,X,ezx)              % Plots nearly indistinguishable

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Source: OpenStax, Applied probability. OpenStax CNX. Aug 31, 2009 Download for free at http://cnx.org/content/col10708/1.6

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