10.1 Function of random vectors (Page 6/6)

Applied probability Page 6 / 6

This graph contains a square that is intersected by a curve. This curve is labeled u=v/t. — Product of $X, Y$ with uniform joint distribution on the unit square.

${X, Y} \sim$ uniform on unit square

$f_{X Y} (t, u) = 1, 0 \leq t \leq 1, 0 \leq u \leq 1$ . Then (see [link] )

P (X Y \leq v) = \int \int_{Q_{v}} 1 d u d t where Q_{v} = {(t, u) : 0 \leq t \leq 1, 0 \leq u \leq min {1, v / t}}

Integration shows

F_{Z} (v) = P (X Y \leq v) = v (1 - ln (v)) so that f_{Z} (v) = - ln (v) = ln (1 / v), 0 < v \leq 1

For $v = 0.5, F_{Z} (0.5) = 0.8466$ .

% Note that although f = 1, it must be expressed in terms of t, u.
tuapprEnter matrix [a b] of X-range endpoints  [0 1]Enter matrix [c d] of Y-range endpoints  [0 1]Enter number of X approximation points  200
Enter number of Y approximation points  200Enter expression for joint density  (u>=0)&(t>=0)
Use array operations on X, Y, PX, PY, t, u, and PG = t.*u;

[Z,PZ] = csort(G,P);p = (Z<=0.5)*PZ'
p =  0.8465                 % Theoretical value 0.8466, above

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Continuation of Example 5 From "random vectors and joint distributions"

The pair ${X, Y}$ has joint density $f_{X Y} (t, u) = \frac{6}{37} (t + 2 u)$ on the region bounded by $t = 0$ , $t = 2$ , $u = 0$ , and $u = max {1, t}$ (see Figure 7). Let $Z = X Y$ . Determine $P (Z \leq 1)$ .

A graph of f_xy(t,u)=(6/37)(t+2u) — Area of integration for [link] .

ANALYTIC SOLUTION

P (Z \leq 1) = P ((X, Y) \in Q) where Q = {(t, u) : u \leq 1 / t}

Reference to [link] shows that

P ((X, Y) \in Q) = \frac{6}{37} \int_{0}^{1} \int_{0}^{1} (t + 2 u) d u d t + \frac{6}{37} \int_{1}^{2} \int_{0}^{1 / t} (t + 2 u) d u d t = 9 / 37 + 9 / 37 = 18 / 37 \approx 0.4865

APPROXIMATE SOLUTION

tuappr
Enter matrix [a b]of X-range endpoints  [0 2]
Enter matrix [c d]of Y-range endpoints  [0 2]
Enter number of X approximation points  300Enter number of Y approximation points  300
Enter expression for joint density  (6/37)*(t + 2*u).*(u<=max(t,1))
Use array operations on X, Y, PX, PY, t, u, and PQ = t.*u<=1;
PQ = total(Q.*P)PQ =  0.4853             % Theoretical value 0.4865, above
G = t.*u;                % Alternate, using the distribution for Z[Z,PZ] = csort(G,P);PZ1 = (Z<=1)*PZ'
PZ1 = 0.4853

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In the following example, the function g has a compound definition. That is, it has a different rule for different parts of the plane.

This is a graph of a curve. To the left of the curve is a shaded area labeled Q_B. Below the curve is a shaded area labeled Q_A. The line defining area Q_B is labeled u=1/2-t. The curve is labeled u=t^2. The top of area Q_A is labeled u=1/2. Inside area Q_A is another triangle labled t_1 and an area labeled t_2 — Regions for $P (Z \leq 1 / 2)$ in [link] .

A compound function

The pair ${X, Y}$ has joint density $f_{X Y} (t, u) = \frac{2}{3} (t + 2 u)$ on the unit square $0 \leq t \leq 1, 0 \leq u \leq 1$ .

Z = \{\begin{matrix} Y & for X^{2} - Y \geq 0 \\ X + Y & for X^{2} - Y < 0 \end{matrix}) = I_{Q} (X, Y) Y + I_{Q^{c}} (X, Y) (X + Y)

for $Q = {(t, u) : u \leq t^{2}}$ . Determine $P (Z < = 0.5)$ .

ANALYTICAL SOLUTION

P (Z \leq 1 / 2) = P (Y \leq 1 / 2, Y \leq X^{2}) + P (X + Y \leq 1 / 2, Y > X^{2}) = P ((X, Y) \in Q_{A} ⋁ Q_{B})

where $Q_{A} = {(t, u) : u \leq 1 / 2, u \leq t^{2}}$ and $Q_{B} = {(t, u) : t + u \leq 1 / 2, u > t^{2}}$ . Reference to [link] shows that this is the part of the unit square for which $u \leq min (max (1 / 2 - t, t^{2}), 1 / 2)$ . We may break up the integral into three parts. Let $1 / 2 - t_{1} = t_{1}^{2}$ and $t_{2}^{2} = 1 / 2$ . Then

P (Z \leq 1 / 2) = \frac{2}{3} \int_{0}^{t_{1}} \int_{0}^{1 / 2 - t} (t + 2 u) d u d t + \frac{2}{3} \int_{t_{1}}^{t_{2}} \int_{0}^{t^{2}} (t + 2 u) d u d t + \frac{2}{3} \int_{t_{2}}^{1} \int_{0}^{1 / 2} (t + 2 u) d u d t = 0.2322

APPROXIMATE SOLUTION

tuappr
Enter matrix [a b]of X-range endpoints  [0 1]
Enter matrix [c d]of Y-range endpoints  [0 1]
Enter number of X approximation points  200Enter number of Y approximation points  200
Enter expression for joint density  (2/3)*(t + 2*u)Use array operations on X, Y, PX, PY, t, u, and P
Q = u<= t.^2;
G = u.*Q + (t + u).*(1-Q);prob = total((G<=1/2).*P)
prob =  0.2328          % Theoretical is 0.2322, above

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The setup of the integrals involves careful attention to the geometry of the system. Once set up, the evaluation is elementary but tedious. On the other hand, the approximationproceeds in a straightforward manner from the normal description of the problem. The numerical result compares quite closely with the theoretical value and accuracy could beimproved by taking more subdivision points.

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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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