10.1 Function of random vectors  (Page 2/6)

The density for the sum $X+Y$

Suppose the pair $\{X,Y\}$ has joint density $f_{XY}$ . Determine the density for

SOLUTION

For any fixed v , the region Q _v is the portion of the plane on or below the line $u=v-t$ (see [link] ). Thus

Differentiating with the aid of the fundamental theorem of calculus, we get

This integral expresssion is known as a convolution integral .

Sum of joint uniform random variables

Suppose the pair $\{X,Y\}$ has joint uniform density on the unit square $0\le t\le 1$ , $0\le u\le 1$ . Determine the density for $Z=X+Y$ .

SOLUTION

$F_Z\left(v\right)$ is the probability in the region $Q_v:u\le v-t$ . Now $P_{XY}\left(Q_v\right)=1-P_{XY}\left(Q_v^c\right)$ , where the complementary set Q _v ^c is the set of points above the line. As Figure 3 shows, for $v\le 1$ , the part of Q _v which has probability mass is the lower shaded triangular region on the figure, which has area (and hence probability) $v^2/2$ . For $v>1$ , the complementary region Q _v ^c is the upper shaded region. It has area ${(2-v)}^2/2$ . so that in this case,

$P_{XY}\left(Q_v\right)=1-{(2-v)}^2/2$ . Thus,

Differentiation shows that Z has the symmetric triangular distribution on $[0,2]$ , since

With the use of indicator functions, these may be combined into a single expression

ALTERNATE SOLUTION

Since $f_{XY}(t,u)=I_{[0,1]}\left(t\right)I_{[0,1]}\left(u\right)$ , we have $f_{XY}(t,v-t)=I_{[0,1]}\left(t\right)I_{[0,1]}(v-t)$ . Now $0\le v-t\le 1$ iff $v-1\le t\le v$ , so that

Integration with respect to t gives the result above.

Independence of simple approximations to an independent pair

Suppose $\{X,Y\}$ is an independent pair with simple approximations X _s and Y _s as described in Distribution Approximations.

As functions of X and Y , respectively, the pair $\{X_s,Y_s\}$ is independent. Also each pair $\{I_{M_i}\left(X\right),I_{N_j}\left(Y\right)\}$ is independent.