Help me fast.... 12x?, 0 The joint density function of the random variables X and Y is given to the right.(a) Show that X and Y are not independent.(b) Find P(X>0.4 | Y = 0.3).(a) Select the correct choice below and fill in the answer box to complete your choice.f(x.y)Since f(xly) =h(y)А.for 0 0.4 |Y=0.3) =D(Simplify your answer. Round to six decimal places as needed.)

Answered: 12x2, 0

Elementary Algebra

17th Edition

ISBN:9780998625713

Author:Lynn Marecek, MaryAnne Anthony-Smith

Publisher:Lynn Marecek, MaryAnne Anthony-Smith

Chapter6: Polynomials

Section6.1: Add And Subtract Polynomials

Problem 6.5TI: Add: 12q2+9q2 .

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Help me fast....

12x?, 0<x<1, 0 <y<1-x
f(x.y) =
0,
elsewhere

The joint density function of the random variables X and Y is given to the right.
(a) Show that X and Y are not independent.
(b) Find P(X>0.4 | Y = 0.3).
(a) Select the correct choice below and fill in the answer box to complete your choice.
f(x.y)
Since f(xly) =
h(y)
А.
for 0 <y<1-x, involves the variable x, X and Y are not independent.
f(x.y)
Since f(xly) =
h(y)
В.
for 0<x<1-y, involves the variable y. X and Y are not independent
OC.
f(x.y)
%3D
Since f(xly) =
for 0 <x<1-y, is constant, X and Y are not independent.
h(y)
D.
f(x.y)
Since f(xly) =
h(y)
for 0 <x<1-y, is a function of only the variable x, X andY are not independent.
(b) P(X>0.4 |Y=0.3) =D
(Simplify your answer. Round to six decimal places as needed.)

Expert Solution

Step 1

(a) The variable X and Y will be independent if the equality $f (x, y) = g (x) h (y)$ holds for all values x and y which the variables are able to assume.

In order to check the independence of the variables, let us find their marginal densities.

$\begin{array}{rcl} g (x) & = & \int_{y} f (x, y) d y \\ = & \int_{0}^{1 - x} 12 x^{2} d y \\ = & 12 x^{2} y |_{0}^{1 - x} \\ = & 12 x^{2} ((1 - x) - 0) \\ = & 12 x^{2} (1 - x), f o r 0 < x < 1 \end{array}$

Further on, we need to find $h (y)$ . For some arbitrary $y \in < 0, 1 >$ because of $y < 1 - x$ we also have:

$x < 1 - y \Rightarrow 0 < x < 1 - y$

Thus, similarly as for $g (x)$ , we obtain:

$\begin{array}{rcl} h (y) & = & \int_{x} f (x, y) d x \\ = & \int_{0}^{1 - y} 12 x^{2} d x \\ = & 12 \int_{0}^{1 - y} x^{2} d x \\ = & 12 . \frac{x^{2 + 1}}{2 + 1} \\ = & 12 . \frac{x^{3}}{3} \\ = & 4 x^{3} |_{0}^{1 - y} \\ = & 4 ({(1 - y)}^{3} - 0^{3}) \\ = & 4 {(1 - y)}^{3}, f o r 0 < y < 1 \end{array}$

Now, we see that $g (x) h (y) = 12 x^{2} (1 - x) . 4 {(1 - y)}^{3}$ which generally isn't equal to $f (x, y) = 12 x^{2}$ . Thus, the random variable X and Y aren't independent.

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