Question

Let Y be the rate (calls per hour) at which calls arrive at a switchboard. Let X be the number of calls during a two-hour period. Suppose that the marginal p.d.f. of Y is

\({{\bf{f}}_{\bf{2}}}\left( {\bf{y}} \right){\bf{ = }}\left\{ {\begin{align}{}{{{\bf{e}}^{{\bf{ - y}}}}}&{{\bf{if}}\,{\bf{y > 0,}}}\\{\bf{0}}&{{\bf{otherwise,}}}\end{align}} \right.\)

And that the conditional p.d.f. of X given\({\bf{Y = y}}\)is

\({{\bf{g}}_{\bf{1}}}\left( {{\bf{x}}\left| {\bf{y}} \right.} \right){\bf{ = }}\left\{ {\begin{align}{}{\frac{{{{\left( {{\bf{2y}}} \right)}^{\bf{x}}}}}{{{\bf{x!}}}}{{\bf{e}}^{{\bf{ - 2y}}}}}&{{\bf{if}}\,{\bf{x = 0,1,}}...{\bf{,}}}\\{\bf{0}}&{{\bf{otherwise}}{\bf{.}}}\end{align}} \right.\)

1. Find the marginal p.d.f. of X. (You may use the formula\(\int_{\bf{0}}^\infty {{{\bf{y}}^{\bf{k}}}{{\bf{e}}^{{\bf{ - y}}}}{\bf{dy = k!}}} {\bf{.}}\))
2. Find the conditional p.d.f.\({{\bf{g}}_{\bf{2}}}\left( {{\bf{y}}\left| {\bf{0}} \right.} \right)\)of Y given\({\bf{X = 0}}\).
3. Find the conditional p.d.f.\({{\bf{g}}_{\bf{2}}}\left( {{\bf{y}}\left| 1 \right.} \right)\)of Y given\({\bf{X = 1}}\).
4. For what values of y is\({{\bf{g}}_{\bf{2}}}\left( {{\bf{y}}\left| {\bf{1}} \right.} \right){\bf{ > }}{{\bf{g}}_{\bf{2}}}\left( {{\bf{y}}\left| {\bf{0}} \right.} \right)\)? Does this agree with the intuition that the more calls you see, the higher you should think the rate is?

Short answer

1. The marginal p.d.f. of X is, \(\frac{1}{3}{\left( {\frac{2}{3}} \right)^x}\)
2. The conditional p.d.f. of \({g_2}\left( {y\left| 0 \right.} \right)\) is, \({g_2}\left( {y\left| 0 \right.} \right) = \left\{ {\begin{align}{}{3{e^{ - 3y}}}&{if\,y > 0}\\0&{otherwise}\end{align}} \right.\)
3. The conditional p.d.f. of \({g_2}\left( {y\left| 1 \right.} \right)\) is, \({g_2}\left( {y\left| x \right.} \right) = \)\(\left\{ {\begin{align}{}{\frac{{\frac{{{{\left( {2y} \right)}^x}}}{{x!}}{e^{ - 3y}}}}{{\frac{1}{2}{{\left( {\frac{2}{3}} \right)}^x}}}}&{if\,y > 0,\,x = 0,1,...}\\0&{otherwise}\end{align}} \right.\)
4. \({g_2}\left( {y\left| 1 \right.} \right) > {g_2}\left( {y\left| 0 \right.} \right)\) for \(y > \frac{1}{3}\)

Step-by-step solution

Given information

The given question deals with the study of the probability density function, along with the conditional probability density function, the joint probability density function, and their relationship.

The provided information in the question is regarding the probability density function, and the conditional probability density function. The random variable Y is the number of calls per hour at a service board, while the random variable X is the number of calls in a two-hour stint.

The probability density function of the random variable Y is,

\({f_2}\left( y \right) = \left\{ {\begin{align}{}{{e^{ - y}}}&{if\,\,y > 0}\\0&{otherwise}\end{align}} \right.\)

On the other hand, the conditional density function of\(X\left| {Y = y} \right.\)is given by,

\({g_1}\left( {x\left| y \right.} \right) = \left\{ {\begin{align}{}{\frac{{{{\left( {2y} \right)}^x}}}{{x!}}{e^{ - 2y}}}&{if\,\,x = 0,1,...,}\\0&{otherwise}\end{align}} \right.\)

Calculations

(a)

The marginal probability density function of X can be obtained by integrating the joint pdf of the random variables X and Y with respect to Y, Now, the joint pdf of the random variable X and Y is obtained as shown below,

\(\begin{array}h\left( {x,y} \right) = {f_1}\left( x \right) \times {g_1}\left( {x\left| y \right.} \right)\\ = \left\{ {\begin{array}{}{\frac{{{{\left( {2y} \right)}^x}}}{{x!}}{e^{ - 3y}}}&{if\,y > 0,\,x = 0,1,...}\\0&{otherwise}\end{array}} \right.\end{array}\)

Thus, the marginal pdf of X is,

\(\begin{align}{f_2}\left( x \right) &= \int_0^\infty {\frac{{{{\left( {2y} \right)}^x}}}{{x!}}{e^{ - 3y}}dy} \\ &= \frac{{{2^x}}}{{x!}}\int_0^\infty {{y^x}{e^{ - 3y}}dy} \\ &= \frac{{{2^x}}}{{x!}}\frac{{x!}}{{{3^{x + 1}}}}\\ &= \frac{1}{3}{\left( {\frac{2}{3}} \right)^x},x = 0,1,...,\end{align}\)

(b)

The conditional pdf of Y given X is obtained by the ratio of the joint pdf to the marginal pdf of X, which is obtained as shown below,

\(\begin{array}{g_2}\left( {y\left| x \right.} \right) = \frac{{h\left( {x,y} \right)}}{{{f_2}\left( x \right)}}\\ = \left\{ {\begin{array}{}{\frac{{\frac{{{{\left( {2y} \right)}^x}}}{{x!}}{e^{ - 3y}}}}{{\frac{1}{3}{{\left( {\frac{2}{3}} \right)}^x}}}}&{if\,y > 0,\,x = 0,1,...}\\0&{otherwise}\end{array}} \right.\end{array}\)

Thus, the required quality is,

\({g_2}\left( {y\left| 0 \right.} \right) = \left\{ {\begin{align}{}{3{e^{ - 3y}}}&{if\,y > 0}\\0&{otherwise}\end{align}} \right.\)

(c)

The conditional pdf of Y given X is obtained by the joint pdf to the marginal pdf of X, which is obtained as shown below,

\(\begin{array}{g_2}\left( {y\left| x \right.} \right) = \frac{{h\left( {y\left| x \right.} \right)}}{{{f_2}\left( x \right)}}\\ = \left\{ {\begin{array}{}{\frac{{\frac{{{{\left( {2y} \right)}^x}}}{{x!}}{e^{ - 3y}}}}{{\frac{1}{2}{{\left( {\frac{2}{3}} \right)}^x}}}}&{if\,y > 0,\,x = 0,1,...}\\0&{otherwise}\end{array}} \right.\end{array}\)

(d)

The given condition that is to be obtained for the value of y is,

\({g_2}\left( {y\left| 1 \right.} \right) > {g_2}\left( {y\left| 0 \right.} \right)\)

The required inequality as shown above can be obtained by the help of the provided inequality in the manner as shown below,

\(\begin{align}{c}9y{e^{ - 3y}} > y{e^{ - 3y}}\\ \Rightarrow y > \frac{1}{3}\end{align}\)

Thus, above agrees with the intuition that, the more calls one see, the more is the rate that is obtained henceforth. The more calls that are received will imply that the rate of the calls in the given interval is large, since the rate is nothing but the number of calls that are received in a particular time period. Hence as the number of calls that are received increases, implies that the rate of the calls will also increase.