## Related questions with answers

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 $f(x,y) = k(x^2+y^2)$, for 30 $\leq x \leq 50$, 30 $\leq y \leq 50,$ f(x, y)=0, elsewhere. Find the covariance of the random variables X and Y.

Solution

VerifiedWe found in the Exercise 3.44, that the $\text{\underline{continuous}}$ random variables $X$ and $Y$ are given with the joint density function

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

We will find the $covariance$ of these variables by using the formula

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

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

$\begin{align*} g(x) &= \int_y f(x,y) \\ &= \int_{30}^{50} \frac{3}{3920000}(x^2+y^2) dy \\ &= \biggr( \frac{3}{3920000}(x^2y+\frac{y^3}{3}) \biggr) \biggr |_{30}^{50} \\ &= \frac{3}{3920000} \bigr( 50x^2+\frac{125000}{3} \bigr) - \frac{3}{3920000}\bigr( 30x^2+9000 \bigr) \\ &= \frac{3x^2}{196000} + \frac{1}{40} \end{align*}$

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

$\begin{align*} h(y) &= \int_x f(x,y) \\ &= \int_{30}^{50} \frac{3}{3920000}(x^2+y^2) dx \\ &= \biggr( \frac{3}{3920000}(xy^2+\frac{x^3}{3}) \biggr) \biggr |_{30}^{50} \\ &= \frac{3}{3920000} \bigr( 50y^2+\frac{125000}{3} \bigr) - \frac{3}{3920000}\bigr( 30y^2+9000 \bigr) \\ &= \frac{3y^2}{196000} + \frac{1}{40} \end{align*}$

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