11.1 Mathematical expectation; general random variables (Page 7/7)

Applied probability Page 7 / 7

A function with a compound definition

The pair ${X, Y}$ has joint density $f_{X Y} (t, u) = 1 / 2$ on the square region bounded by $u = 1 + t$ , $u = 1 - t$ , $u = 3 - t$ , and $u = t - 1$ (see [link] ),

W = \{\begin{matrix} X & for max {X, Y} \leq 1 \\ 2 Y & for max {X, Y} > 1 \end{matrix}) = I_{Q} (X, Y) X + I_{Q^{c}} (X, Y) 2 Y

where $Q = {(t, u) : max {t, u} \leq 1} = {(t, u) : t \leq 1, u \leq 1}$ . Determine $E [W]$ .

Figure 2 is a density drawing, with a horizontal axis labeled as t, and a vertical axis labeled u. The drawing is a shaded square rotated 45 degrees to be sitting with one point on the horizontal axis. The point sits on (1, 0) and a second point sits against the vertical axis, at (0, 1). In looking at the drawing it can be deduced that the third vertex is at (1, 2), and that the fourth vertex is at (2, 1). Each side of the square is labeled with an equation. Starting with the side between the vertices that are sitting on the axes, an reading them clockwise, the equations are listed as u= 1 - t, u = 1 + t, u= 3 - t, and u = t - 1. There is also an equation inside the shaded square, reading f_xy (t, u) = 1/2. — The density for [link] .

ANALYTIC SOLUTION

The intersection of the region Q and the square is the set for which $0 \leq t \leq 1$ and $1 - t \leq u \leq 1$ . Reference to the figure shows three regions of integration.

E [W] = \frac{1}{2} \int_{0}^{1} \int_{1 - t}^{1} t d u d t + \frac{1}{2} \int_{0}^{1} \int_{1}^{1 + t} 2 u d u d t + \frac{1}{2} \int_{1}^{2} \int_{t - 1}^{3 - t} 2 u d u d t = 11 / 6 \approx 1.8333

APPROXIMATION

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  200Enter number of Y approximation points  200
Enter expression for joint density  ((u<=min(t+1,3-t))&...
      (u>=max(1-t,t-1)))/2
Use array operations on X, Y, PX, PY, t, u, and PM = max(t,u)<=1;
G = t.*M + 2*u.*(1 - M);   % Z = g(X,Y)EG = total(G.*P)           % E[g(X,Y)]
EG =  1.8340               % Theoretical 11/6 = 1.8333[Z,PZ] = csort(G,P);       % Distribution for ZEZ = dot(Z,PZ)             % E[Z] from distributionEZ =  1.8340

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Special forms for expectation

The various special forms related to property (E20a) are often useful. The general result, which we do not need, is usually derived by an argument which employs a generalform of what is known as Fubini's theorem. The special form (E20b)

E [X] = \int_{- \infty}^{\infty} [u (t) - F_{X} (t)] d t

may be derived from (E20a) by use of integration by parts for Stieltjes integrals. However, we use the relationship between the graph of the distribution function and the graph ofthe quantile function to show the equivalence of (E20b) and (E20f) . The latter property is readily established by elementary arguments.

The property (E20f)

If Q is the quantile function for the distribution function F _X , then

E [g (X)] = \int_{0}^{1} g [Q (u)] d u

VERIFICATION

If $Y = Q (U)$ , where $U \sim$ uniform on $(0, 1)$ , then Y has the same distribution as X . Hence,

E [g (X)] = E [g (Q (U))] = \int g (Q (u)) f_{U} (u) d u = \int_{0}^{1} g (Q (u)) d u

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Reliability and expectation

In reliability, if X is the life duration (time to failure) for a device, the reliability function is the probability at any time t the device is still operative. Thus

R (t) = P (X > t) = 1 - F_{X} (t)

According to property (E20b)

E [X] = \int_{0}^{\infty} R (t) d t

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Use of the quantile function

Suppose $F_{X} (t) = t^{a}, a > 0, 0 \leq t \leq 1$ . Then $Q (u) = u^{1 / a}, 0 \leq u \leq a$ .

E [X] = \int_{0}^{1} u^{1 / a} d u = \frac{1}{1 + 1 / a} = \frac{a}{a + 1}

The same result could be obtained by using $f_{X} (t) = F_{X}^{'} (t)$ and evaluating $\int t f_{X} (t) d t$ .

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Equivalence of (E20b) And (E20f)

For the special case $g (X) = X$ , Figure 3(a) shows $\int_{0}^{1} Q (u) d u$ is the difference in the shaded areas

\int_{0}^{1} Q (u) d u = Area A - Area B

The corresponding graph of the distribution function F is shown in Figure 3(b) . Because of the construction, the areas of the regions marked A and B are the same in the two figures. As may be seen,

Area A = \int_{0}^{\infty} [1 - F (t)] d t and Area B = \int_{- \infty}^{0} F (t) d t

Use of the unit step function $u (t) = 1$ for $t > 0$ and 0 for $t < 0$ (defined arbitrarily at $t = 0$ ) enables us to combine the two expressions to get

\int_{0}^{1} Q (u) d u = Area A - Area B = \int_{- \infty}^{\infty} [u (t) - F (t)] d t

Figure three contains two graphs. The first graph has a horizontal axis labeled t, and a vertical axis labeled u. The large label of the graph reads, u = Q(t). A dashed vertical line along t = 1 bounds an increasing curved plot. The curve starts with a vertical asymptote along the vertical axis below the horizontal axis, and as it approaches the horizontal axis, the slope becomes more shallow. The curve's slope shallows until it is midway in horizontal distance between the vertical axis and the dashed vertical line. At this point, the slope begins to increase again, until it reaches a vertical asymptote along the dashed line at t = 1. The horizontal and vertical axes, along with the curve itself, create a bounded shape. A small right triangle loosely fits this bounded shape, and is labeled as B. The dashed line, horizontal axis, and the segment of the curve above the horizontal axis create a larger bounded shape, and a larger right triangle loosely fits this bounded shape, labeled A. The second graph is roughly similar. The axes are in the same place, but with this figure, s dashed line is now drawn horizontally along u = 1. A curve of the same shape now begins as a horizontal asymptote along the t - axis. It increases in slope at an increasing rate for half of the vertical distance and then decreases in slope back to a horizontal asymptote at u = 1. The same triangles fitting the same bounded regions as in the first figure are used in the second figures, only because of the rotated nature of the new curve, these triangles are rotated in the same fashion. — Equivalence of properties (E20b) and (E20f) .

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Property (E20c) is a direct result of linearity and (E20b) , with the unit step functions cancelling out.

Property (E20d) Useful inequalities

Suppose $X \geq 0$ . Then

\sum_{n = 0}^{\infty} P (X \geq n + 1) \leq E [X] \leq \sum_{n = 0}^{\infty} P (X \geq n) \leq N \sum_{k = 0}^{\infty} P (X \geq k N), for all N \geq 1

VERIFICATION

For $X \geq 0$ , by (E20b)

E [X] = \int_{0}^{\infty} [1 - F (t)] d t = \int_{0}^{\infty} P (X > t) d t

Since F can have only a countable number of jumps on any interval and $P (X > t)$ and $P (X \geq t)$ differ only at jump points, we may assert

\int_{a}^{b} P (X > t) d t = \int_{a}^{b} P (X \geq t) d t

For each nonnegative integer n , let $E_{n} = [n, n + 1)$ . By the countable additivity of expectation

E [X] = \sum_{n = 0}^{\infty} E [I_{E_{n}} X] = \sum_{n = 0}^{\infty} \int_{E_{n}} P (X \geq t) d t

Since $P (X \geq t)$ is decreasing with t and each E _n has unit length, we have by the mean value theorem

P (X \geq n + 1) \leq E [I_{E_{n}} X] \leq P (X \geq n)

The third inequality follows from the fact that

\int_{k N}^{(k + 1) N} P (X \geq t) d t \leq N \int_{E_{k N}} P (X \geq t) d t \leq N P (X \geq k N)

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Remark . Property (E20d) is used primarily for theoretical purposes. The special case (E20e) is more frequently used.

Property (E20e)

If X is nonnegative, integer valued, then

E [X] = \sum_{k = 1}^{\infty} P (X \geq k) = \sum_{k = 0}^{\infty} P (X > k)

VERIFICATION

The result follows as a special case of (E20d) . For integer valued random variables,

P (X \geq t) = P (X \geq n) on E_{n} and P (X \geq t) = P (X > n) = P (X \geq n + 1) on E_{n + 1}

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An elementary derivation of (E20e) can be constructed as follows.

(E20e) For integer-valued random variables

By definition

E [X] = \sum_{k = 1}^{\infty} k P (X = k) = lim_{n} \sum_{k = 1}^{n} k P (X = k)

Now for each finite n ,

\sum_{k = 1}^{n} k P (X = k) = \sum_{k = 1}^{n} \sum_{j = 1}^{k} P (X = k) = \sum_{j = 1}^{n} \sum_{k = j}^{n} P (X = k) = \sum_{j = 1}^{n} P (X \geq j)

Taking limits as $n \to \infty$ yields the desired result.

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The geometric distribution

Suppose $X \sim$ geometric $(p)$ . Then $P (X \geq k) = q^{k}$ . Use of (E20e) gives

E [X] = \sum_{k = 1}^{\infty} q^{k} = q \sum_{k = 0}^{\infty} q^{k} = \frac{q}{1 - q} = q / p

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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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