14.1 Problems on conditional expectation, regression (Page 4/5)

Applied probability Page 4 / 5

$f_{X Y} (t, u) = \frac{3}{23} (t + 2 u)$ for $0 \leq t \leq 2$ , $0 \leq u \leq max {2 - t, t}$ (see Exercise 39 , and [link] ).

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

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

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

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

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

= I_{[0, 1]} (t) \frac{1}{12} \cdot \frac{2 t^{3} - 30 t^{2} + 69 t - 60}{t - 2} + I_{(1, 2]} (t) \frac{7}{6} 2 t

% Continuation of
[link] M = X<= 1;
Q = (t<=1)&(u<=1);
G = (t+u).*Q + 2*u.*(1-Q);EZx = sum(G.*P)./sum(P);
ezx = (1/12)*M.*(2*X.^3 - 30*X.^2 + 69*X -60)./(X-2) + (7/6)*X.*(1-M);plot(X,EZx,X,ezx)

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$f_{X Y} (t, u) = \frac{2}{13} (t + 2 u)$ , for $0 \leq t \leq 2$ , $0 \leq u \leq min {2 t, 3 - t}$ . (see Exercise 31 from "Problems on Mathematical Expectation", and [link] ).

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

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

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

I_{M} (t, u) = I_{[0, 1]} (t) I_{[1, 2]} (u) I_{M^{c}} (t, u) = I_{[0, 1]} (t) I_{[0, 1)} (u) + I_{(1, 2]} (t) I_{[0, 3 - t]} (u)

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

I_{(1 / 2, 1]} (t) [\frac{1}{6 t^{2}} \int_{0}^{1} 2 u^{2} (t + 2 u) d u + \frac{1}{6 t^{2}} \int_{1}^{2 t} (t + u) (t + 2 u) d u] + I_{(1, 2]} (t) \frac{1}{3 (3 - t)} \int_{0}^{3 - t} 2 u^{2} (t + 2 u) d u

= I_{[0, 1 / 2]} (t) \frac{32}{9} t^{2} + I_{(1 / 2, 1]} (t) \frac{1}{36} \cdot \frac{80 t^{3} - 6 t^{2} - 5 t + 2}{t^{2}} + I_{(1, 2]} (t) \frac{1}{9} (- t^{3} + 15 t^{2} - 63 t + 81)

tuappr:  [0 2] [0 2]200 200 (2/13)*(t + 2*u).*(u<=min(2*t,3-t))
M = (t<=1)&(u>=1);
Q = (t+u).*M + 2*(1-M).*u.^2;EZx = sum(Q.*P)./sum(P);
N1 = X<= 1/2;
N2 = (X>1/2)&(X<=1);
N3 = X>1;
ezx = (32/9)*N1.*X.^2 + (1/36)*N2.*(80*X.^3 - 6*X.^2 - 5*X + 2)./X.^2 ...+ (1/9)*N3.*(-X.^3 + 15*X.^2 - 63.*X + 81);
plot(X,EZx,X,ezx)

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$f_{X Y} (t, u) = I_{[0, 1]} (t) \frac{3}{8} (t^{2} + 2 u) + I_{(1, 2]} (t) \frac{9}{14} t^{2} u^{2}$ , for $0 \leq u \leq 1$ . (see Exercise 32 from "Problems on Mathematical Expectation", and [link] ).

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

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

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

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

= I_{[0, 1]} (t) t + I_{(1, 2]} (t) (- \frac{13}{4} t + 12 t^{2} - 12 t^{3} + 5 t^{4} - \frac{3}{4} t^{5})

tuappr:  [0 2] [0 1]200 100  (t<=1).*(t.^2 + 2*u)./(t.^2 + 1) +3*u.^2.*(t>1)
M = u<=min(1,2-t);
G = M.*t + (1-M).*t.*u;EZx = sum(G.*P)./sum(P);
N = X<=1;
ezx = X.*N + (1-N).*(-(13/4)*X + 12*X.^2 - 12*X.^3 + 5*X.^4 - (3/4)*X.^5);plot(X,EZx,X,ezx)

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Suppose $X \sim$ uniform on 0 through n and $Y \sim$ conditionally uniform on 0 through i , given $X = i$ .

Determine $E [Y]$ from $E [Y | X = i]$ .
Determine the joint distribution for ${X, Y}$ for $n = 50$ (see Example 7 from "Conditional Expectation, Regression" for a possible approach). Use jcalc to determine $E [Y]$ ; compare with the theoretical value.

$E [Y | X = i] = i / 2$ , so
$E [Y] = \sum_{i = 0}^{n} E [Y | X = i] P (X = i) = \frac{1}{n + 1} \sum_{i = 1}^{n} i / 2 = n / 4$
$P (X = i) = 1 / (n + 1), 0 \leq i \leq n$ , $P (Y = k | X = i) = 1 / (i + 1), 0 \leq k \leq i$ ; hence $P (X = i, Y = k) = 1 / (n + 1) (i + 1) 0 \leq i \leq n, 0 \leq k \leq i$ . n = 50; X = 0:n; Y = 0:n; P0 = zeros(n+1,n+1);for i = 0:n P0(i+1,1:i+1) = (1/((n+1)*(i+1)))*ones(1,i+1);end P = rot90(P0);jcalc: X Y P - - - - - - - - - - -EY = dot(Y,PY) EY = 12.5000 % Comparison with part (a): 50/4 = 12.5

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Suppose $X \sim$ uniform on 1 through n and $Y \sim$ conditionally uniform on 1 through i , given $X = i$ .

Determine $E [Y]$ from $E [Y | X = i]$ .
Determine the joint distribution for ${X, Y}$ for $n = 50$ (see Example 7 from "Conditional Expectation, Regression" for a possible approach). Use jcalc to determine $E [Y]$ ; compare with the theoretical value.

$E [Y | X = i] = (i + 1) / 2$ , so
$E [Y] = \sum_{i = 1}^{n} E [Y | X = i] P (X = i) = \frac{1}{n} \sum_{i = 1}^{n} \frac{i + 1}{2} = \frac{n + 3}{4}$
$P (X = i) = 1 / n, 1 \leq i \leq n$ , $P (Y = k | X = i) = 1 / i, 1 \leq k \leq i$ ; hence $P (X = i, Y = k) = 1 / n i 1 \leq i \leq n, 1 \leq k \leq i$ . n = 50; X = 1:n; Y = 1:n; P0 = zeros(n,n);for i = 1:n P0(i,1:i) = (1/(n*i))*ones(1,i);end P = rot90(P0);jcalc: P X Y - - - - - - - - - - - -EY = dot(Y,PY) EY = 13.2500 % Comparison with part (a): 53/4 = 13.25

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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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