Question

Question:LetYbe the rate (calls per hour) at which calls arrive at a switchboard. LetXbe the number of calls during at wo-hour period. A popular choice of joint p.f./p.d.f. for(X, Y )in this example would be one like

\({\bf{f}}\left( {{\bf{x,y}}} \right){\bf{ = }}\left\{ \begin{array}{l}\frac{{{{\left( {{\bf{2y}}} \right)}^{\bf{x}}}}}{{{\bf{x!}}}}{{\bf{e}}^{{\bf{ - 3y}}}}\;{\bf{if}}\;{\bf{y > 0}}\;{\bf{and}}\;{\bf{x = 0,1, \ldots }}\\{\bf{0}}\;{\bf{otherwise}}\end{array} \right.\)

a. Verify thatfis a joint p.f./p.d.f. Hint:First, sum overthexvalues using the well-known formula for thepower series expansion of\({{\bf{e}}^{{\bf{2y}}}}\).

b. Find Pr(X=0).

Short answer

1. It is verified that f is a joint p.f/p.d.f.
2. \(\Pr \left( {X = 0} \right) = \frac{1}{3}\)

Step-by-step solution

Given information

At a switchboard, the rate which is arriving a call per hour is Y and for a two-hour period, the number of calls is X.

The joint probability density function of (X,Y) is,

\(f\left( {x,y} \right) = \left\{ \begin{array}{l}\frac{{{{\left( {2y} \right)}^x}}}{{x!}}{e^{ - 3y}}\;if\;y > 0\;and\;x = 0,1, \ldots \\0\;otherwise\end{array} \right.\)

Determining the verification

a. .

To verify that f is a joint p.d.f or p.f and to verify its validity, we have to do the sum over all x values and integrate the sum over the region\({{\bf{S}}_{\bf{y}}}{\bf{ = }}\left\{ {{\bf{y:y > 0}}} \right\}\). Then we have to show that the value of the integration is 1.

So,

\(\begin{array}{c}\int\limits_0^\infty {\sum\limits_{x = 0}^\infty {{f_{X,Y}}\left( {x,y} \right)dy = } } \int\limits_0^\infty {\sum\limits_{x = 0}^\infty {\frac{{{{\left( {2y} \right)}^x}}}{{x!}}{e^{ - 3y}}dy} } \\ = \int\limits_0^\infty {{e^{ - 3y}}\sum\limits_{x = 0}^\infty {\frac{{{{\left( {2y} \right)}^x}}}{{x!}}dy} } \\ = \int\limits_0^\infty {{e^{ - 3y}}{e^{2y}}dy} \\ = \left. {{e^{ - y}}} \right|_0^\infty \\ = 1\end{array}\)[For the power series of\({e^{2y}}\)]

Thus, it is verified that f is joint p.d.f.

Calculating the probability

b.

The probability that X=0is-

\(\begin{array}{c}\Pr \left( {X = 0} \right) = {f_X}\left( 0 \right)\\ = \int\limits_0^\infty {{f_{X,Y}}\left( {0,y} \right)dy} \\ = \int\limits_0^\infty {\frac{{{{\left( {2y} \right)}^0}}}{{0!}}{e^{ - 3y}}dy} \\ = \int\limits_0^\infty {{e^{ - 3y}}dy} \\ = \left. {\frac{{{e^{ - 3y}}}}{3}} \right|_0^\infty \\ = \frac{1}{3}\end{array}\)

Thus, the probability is \(\frac{1}{3}\).