Q .7.3 - A First Course in Probability

Question

Prove Proposition $2.1$when

$\left(a\right)$ $X$ and $Y$ have a joint probability mass function;

$\left(b\right) X$ and $Y$ have a joint probability density function and

$g(x,y)\ge 0$ for all $x,y$.

Step-by-Step Solution

Answer

(a). A joint probability mass function is proved as $\sum _{z\in g(A,B)}zP\left(g\right(X,Y)=z)=\sum _{x,y}g(x,y)p(x,y).$

(b). A joint probability density function is proved as ${\int }_0^{\infty }{\iint }_{g(x,y)>t}f(x,y)dxdydt={\iint }_{x,y}{\int }_0^{g(x,y)}f(x,y)dtdxdy={\iint }_{x,y}g(x,y)f(x,y)dxdy$.

Step 1: Given Information part(a)

A joint probability function.

Step 2: Explanation part(a)

Hence, the random variable assumes is,

The mean is

$E\left(g\right(X,Y\left)\right)=\sum _{z\in g(A,B)}zP\left(g\right(X,Y)=z)$

Now, observe that every $z\in g(A,B)$ has form $z=g(x, y)$ for some $x in A$ and $y in B$.

Step 3: Explanation part(a)

Also we have that,

$P\left(g\right(X, Y)=z)=P\left(g\right(X, Y)=g(x, y\left)\right)=P\left(\right(X, Y)=(x, y\left)\right)=p(x, y)$

So we finally we have that,

$\sum _{z\in g(A,B)}zP\left(g\right(X,Y)=z)=\sum _{x,y}g(x,y)p(x,y)$

Therefore, we have proved the claimed.

Step 4: Final answer part(a)

A joint probability mass function is proved as $\sum _{z\in g(A,B)}zP\left(g\right(X,Y)=z)=\sum _{x,y}g(x,y)p(x,y)$.

Step 5: Given Information part(b)

A  joint probability density function

Step 6: Explanation part(b)

 Using the fact that the expectation of random variable can be written as an integral where we integrate , we have that

Also we have that,

Step 7: Explanation part(b)

which yields that

Changing the order of integration, we have that

so we have proved the claimed.

Step 8: Final answer part(b)

A joint probability density function is proved as .