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 Determine whether the two random variables X and Y are dependent or independent, if they are given in the following way. \text{ Determine whether the two random variables } X \text{ and } Y \text{ are dependent or independent, if they are given in the following way. }

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 Each rear tire on an experimental airplane is supposed to be filled to a pressure of 40 pounds per square inch (psi). Let X denote the actual air pressure for the right tire and Y denote the actual air pressure for the left tire. Suppose that X and Y are random variables with the joint density function \text{ Each rear tire on an experimental airplane is supposed to be filled to a pressure of 40 pounds per square inch (psi). Let } X \text{ denote the actual air pressure for the right tire and } Y \text{ denote the actual air pressure for the left tire. Suppose that } X \text{ and } Y \text{ are random variables with the joint density function }

f(x,y)=k(x2+y2), for 30x50, 30y50,f(x,y) = k(x^2+y^2), \text{ for } 30 \leq x \leq 50, \text{ } 30 \leq y \leq 50,

f(x,y)=0, elsewhere.f(x,y)=0, \text{ elsewhere.}

Question

Each rear tire on an experimental airplane is supposed to be filled to a pressure of 40 pounds per square inch (psi). Let X denote the actual air pressure for the right tire and Y denote the actual air pressure for the left tire. Suppose that X and Y are random variables with the joint density function f(x,y)=k(x2+y2)f(x,y) = k(x^2+y^2), for 30 x50\leq x \leq 50, 30 y50,\leq y \leq 50, f(x, y)=0, elsewhere. Find the covariance of the random variables X and Y.

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We found in the Exercise 3.44, that the continuous\text{\underline{continuous}} random variables XX and YY are given with the joint density function

f(x,y)={33920000(x2+y2),30x<50, 30y<500,elsewheref(x,y)= \begin{cases} \frac{3}{3920000}(x^2+y^2), & 30 \leq x < 50, \text{ } 30 \leq y < 50\\ 0, & \text{elsewhere} \end{cases}

We will find the covariancecovariance of these variables by using the formula

σXY=E(XY)μXμY\begin{align} \sigma_{XY} = E(XY) - \mu_X \mu_Y \end{align}

Therefore, we first need to find the marginal distributions\textit{marginal distributions} of XX and YY to be able to find μX\mu_X and μY\mu_Y. Using, g(x)=yf(x,y)g(x) = \int_y f(x,y), we obtain:

g(x)=yf(x,y)=305033920000(x2+y2)dy=(33920000(x2y+y33))3050=33920000(50x2+1250003)33920000(30x2+9000)=3x2196000+140\begin{align*} g(x) &= \int_y f(x,y) \\ &= \int_{30}^{50} \frac{3}{3920000}(x^2+y^2) dy \\ &= \biggr( \frac{3}{3920000}(x^2y+\frac{y^3}{3}) \biggr) \biggr |_{30}^{50} \\ &= \frac{3}{3920000} \bigr( 50x^2+\frac{125000}{3} \bigr) - \frac{3}{3920000}\bigr( 30x^2+9000 \bigr) \\ &= \frac{3x^2}{196000} + \frac{1}{40} \end{align*}

Similarly, by the use of h(y)=xf(x,y)h(y) = \int_x f(x,y), we get:

h(y)=xf(x,y)=305033920000(x2+y2)dx=(33920000(xy2+x33))3050=33920000(50y2+1250003)33920000(30y2+9000)=3y2196000+140\begin{align*} h(y) &= \int_x f(x,y) \\ &= \int_{30}^{50} \frac{3}{3920000}(x^2+y^2) dx \\ &= \biggr( \frac{3}{3920000}(xy^2+\frac{x^3}{3}) \biggr) \biggr |_{30}^{50} \\ &= \frac{3}{3920000} \bigr( 50y^2+\frac{125000}{3} \bigr) - \frac{3}{3920000}\bigr( 30y^2+9000 \bigr) \\ &= \frac{3y^2}{196000} + \frac{1}{40} \end{align*}

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