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Suppose that an automobile dealer pays an amount X (in thousands of dollars) for a used car and then sells it for an amount Y. Suppose that the random variables X and Y have the following joint p.d.f. Determine the dealer’s expected gain from the sale.

\(f\left( {x,y} \right) = \left\{ \begin{aligned}{}\frac{1}{{36}}x\;\;\;\;\;\;\;\;\;\;for0 < x < y < 6\\0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;otherwise.\end{aligned} \right.\)

Short Answer

Expert verified

The dealer’s expected gain is\(\frac{3}{2}\)

Step by step solution

01

Given information

An automobile dealer pays an amount X for a used car and sells it for an amount Y.

02

Calculating the dealer’s expected to gain 

Dealer’s expected gain is given by\(E\left( {Y - X} \right)\)

\(\begin{aligned}{}E\left( {Y - X} \right) = \frac{1}{{36}}\int\limits_0^6 {\int\limits_0^y {\left( {y - x} \right)xdxdy} } \\ = \frac{1}{{36}}\int\limits_0^6 {\frac{{{y^3}}}{6}dy} \\ = \frac{1}{{216}}\int\limits_0^6 {{y^3}dy} \end{aligned}\)

\(\begin{aligned}{}E\left( {Y - X} \right) = \frac{1}{{216}}\left( {\frac{{{y^4}}}{4}} \right)_0^6\\ = \frac{1}{{216}}\frac{{1296}}{4}\\ = \frac{{324}}{{216}}\end{aligned}\)

\(E\left( {Y - X} \right)\)\( = \frac{3}{2}\)

Hence, the dealer’s expected gain is \(\frac{3}{2}\)

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