Question

Let X denote the temperature ( ) and let Y denote the time in minutes that it takes for the diesel engine on an automobile to get ready to start. Assume that the joint density function is
f(x, y) =c (4x+2y+1) , . i) Find c, ii) Find the marginal densities for X and Y, iii) Find the probability that on a randomly selected day the air air temperature will exceed
iv) Are X and Y independent?

Answer 1

Answer on Question #45889 – Math – Statistics and Probability

Let $X$ denote the temperature ( ) and let $Y$ denote the time in minutes that it takes for the diesel engine on an automobile to get ready to start. Assume that the joint density function is $f_{XY}(x,y) = c(4x + 2y + 1)$.

$$0 \leq x \leq 40, \quad 0 \leq y \leq 2$$

i) Find c, ii) Find the marginal densities for X and Y, iii) Find the probability that on a randomly selected day the air temperature will exceed iv) Are X and Y independent?

Solution

i) The integral of the joint distribution over its domain must be 1. So,

$$\int_{0}^{40} dx \int_{0}^{2} dy c(4x + 2y + 1) = \int_{0}^{2} dy c(2x^2 + 2xy + x)(x = 40) - \int_{0}^{2} dy c(2x^2 +$$ $$2xy + x)(x = 0) = \int_{0}^{2} dy c(80y + 3240) = 40c[y^2 + 81y](y = 2) - 40c[y^2 +$$ $$81y](y = 0) = 40c[4 + 162] = 6640c = 1$$ $$c = \frac{1}{6640}.$$

ii) The marginal distribution Fax for $x$ is

$$\begin{aligned} f_{x}(x) = & \int_{0}^{2} f_{XY}(x,y) dy = \frac{1}{6640} \int_{0}^{2} dy(4x + 2y + 1) = \\ & = \frac{1}{6640} [4xy + y^2 + y](y = 2) - \frac{1}{6640} [4xy + y^2 + y](y = 0) \\ & = \frac{1}{6640} (8x + 6). \end{aligned}$$

For $y$, we have

$$\begin{aligned} f_{y}(y) = & \int_{0}^{40} f_{XY}(x,y) dx = \frac{1}{6640} \int_{0}^{40} dx(4x + 2y + 1) = \\ & = \frac{1}{6640} [2x^2 + 2yx + x](x = 40) - \frac{1}{6640} [2x^2 + 2yx + x](x = 0) \\ & = \frac{1}{6640} (80y + 3240). \end{aligned}$$

iii) It's the probability that $X > X_0$ and $Y > Y_0$. So,

$$\begin{aligned} P(X > X_0, Y > Y_0) = & \frac{1}{6640} \int_{X_0}^{40} dx \int_{Y_0}^{2} dy(4x + 2y + 1) \\ & = \frac{4}{6640} \int_{X_0}^{40} x dx \int_{Y_0}^{2} dy + \frac{2}{6640} \int_{X_0}^{40} dx \int_{Y_0}^{2} y dy + \frac{1}{6640} \int_{X_0}^{40} dx \int_{Y_0}^{2} dy = \\ & = \left. \frac{4}{6640} \frac{x^2}{2} \right|_{x = X_0}^{40} y|_{y = Y_0}^2 + \left. \frac{2}{6640} x \right|_{x = X_0}^{40} \frac{y^2}{2} \Bigg|_{y = Y_0}^2 + \frac{1}{6640} x \big|_{x = X_0}^{40} y\big|_{y = Y_0}^2 = \\ & = \frac{1}{3320} (40^2 - X_0^2)(2 - Y_0) + \frac{1}{6640} (40 - X_0)(2^2 - Y_0^2) + \\ \end{aligned}$$

$+\frac{1}{6640} (40 - X_0)(2 - Y_0)$ over the rectangle

$$\{X_0 < x \leq 40, Y_0 \leq y \leq 2\}.$$

iv) No, because their joint distribution is not the product of their marginal distributions.

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