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Chapter 2

Q. 2.11

Let the joint distribution of (X,Y) be such that the possible outcomes (u , v) are the points within the rectangle   (-1 \leq u \leq 2,- 1 \leq v \leq 1) . Furthermore, let every one of these possible outcomes be “equally likely.” Note that we are not dealing with discrete random variables in this example, because the set of possible values is not discrete. As with a single random variable on a continuous set of possible values, the term equally likely denotes a constant value of the probability density function. Thus, we say that


for -1 \leq u \leq 2,- 1 \leq v \leq 1)



in which C is some constant. Find the value of the constant C and the joint cumulative distribution function F_{XY}(u,v) for all values of (u,v) .


Verified Solution

We find the value of C by using the total probability property that (X,Y) must fall somewhere within the space of possible values and that the probability of being within a set is the double integral over that set, as in Eq. 2.21.

P[(X,Y)\in A]=\int_{A}^{}{} \int_{}^{}{p_{XY}(u,v)du dv}

Thus, we obtain C = 1/6, because the rectangle of possible values has an area of 6 and the joint probability density function has the constant value C everywhere in the rectangle:

1=P(-1 \leq X \leq 2,-1 \leq Y \leq 1)= \int_{-1}^{1}{} \int_{-1}^{2}{p_{XY}(u,v)du dv}=6C

We may now calculate the joint cumulative distribution of these two random variables by integrating as in Eq. 2.20,

F_{\vec{X}}(\vec{u})=\int_{-\infty}^{u_{n}}{…} \int_{-\infty}^{u_{1}}{p_{\vec{X}}(\vec{w})dw_{1}…dw_{n}}

with the result that

F_{XY}(u,v)=C(u+1)(v+1)=(u+1)(v+1)/6 for -1 \leq u \leq 2,- 1 \leq v \leq 1
F_{XY}(u,v)=3C(v+1)=(v+1)/2 for u \gt 2,- 1 \leq v \leq 1
F_{XY}(u,v)=2C(u+1)=(u+1)/3 for -1 \leq u \leq 2, v \gt 1
F_{XY}(u,v)=6C=1 for u \gt 2, v \gt 1
F_{XY}(u,v)=0 otherwise

One can verify that this cumulative distribution function is continuous. Thus, the random variables X and Y are said to have a continuous joint distribution. Clearly, for this particular problem the description given by the density function is much simpler than that given by the cumulative distribution function.