Question

Determine the value of c such that the function f(x,y) = cxy for 0 < x < 3 and 0 < y < 3 satisfies the properties of a joint probability density function. Determine the following. Round your answers to four decimal places (e.g. 98.7654).

Answer 1

Answer:

Thus the value of c such that the function f(x,y) = cxy for 0 < x < 3 and 0 < y < 3 satisfies the properties of a joint probability density function is:

c=0.0494

Step-by-step explanation:

Lets start by understanding what we are looking for, thus what is a Joint Probability Density Function.

In principle, and considering a given Probability Space, if two variables exist, say $X$ and $Y$, then the Joint Probability defined as $f_{XY}(x,y)$, essentially denotes the probability that each $X$ and $Y$ exist in a discrete value set,  particularly specified for these variables.

If the given variables, (here $X$ and $Y$) are continuous, then three properties must be met by $f_{XY}(x,y)$. However in this question we are only interested in the first two properties, defined as:

• $f_{XY}(x,y)\geq 0$ for all $x,y$
• $\int\limits^\infty_{-\infty} \int\limits^\infty_{-\infty}{} \,f_{XY}(x,y) dydx=1$

Now lets look at our case here. We know the following:

• $f_{XY}(x,y)=cxy$
• $0<x<3$
• $0<y<3$

Thus we can say that:

$f_{XY}(x,y)=0$,    otherwise

$f_{XY}(x,y)=cxy$,    $0<x<3$, $0<y<3$

With respect to that, then we can write our integral for solution as follow:

$\int\limits^3_b {}\int\limits^3_b {cxy} \, dydx =1$       Eqn(1).

Having obtained Eqn(1) we can start solving for our constant $c$ as follow:

$c\int\limits^3_b {}\int\limits^3_b {xy} \, dydx =1$            Take constant outside of the Integrals

$c\int\limits^3_b{x\frac{y^2}{2} }|^3_0 \, dx =1$             Compute first integral for $dy$

$c\int\limits^3_b{x[\frac{3^2}{2}-\frac{0^2}{2}] } \, dx =1$       Apply Boundary Conditions

$c\int\limits^3_b{\frac{9}{2}x } \, dx =1$                Compute and Simplify

$\frac{9}{2}c[ \frac{x^2}{2}]|^3_0=1$                         Compute second integral for  $dx$

$\frac{9}{2}c[ \frac{3^2}{2}- \frac{0^2}{2}]=1$             Apply Boundary Conditions

$\frac{81}{4}c=1$                          Simplify and solve for c

$81c=4\\ c=\frac{4}{81}\\ c=0.04938$

$c=0.0494$                        Answer rounded to 4 d.p.

Thus the value of c such that the function f(x,y) = cxy for 0 < x < 3 and 0 < y < 3 satisfies the properties of a joint probability density function is:

$c=0.0494$

Answer 2

The value of c such that the function f(x,y) = cxy for 0 < x < 3 and 0 < y < 3 satisfies the properties of a joint probability density function is 0.0494

The probability density function is given as;

f(x,y) = cxy for 0 < x < 3 and 0 < y < 3

For the density function to be a joint probability function, the following must be true

$\int\limits^a_b {\int\limits^a_b {f(x,y)} \, dy } \, dx = 1$

So, we have:

$\int\limits^3_0 {\int\limits^3_0 {cxy} \, dy } \, dx = 1$

Factor out the constant

$c\int\limits^3_0 {\int\limits^3_0 {xy} \, dy } \, dx = 1$

Integrate

$c\int\limits^3_0 x* \frac{y^2}2|\limits^3_0 } \, dx = 1$

Expand

$c\int\limits^3_0 x* \frac{3^2}2} \, dx = 1$

$c\int\limits^3_0 x* \frac{9}2} \, dx = 1$

Factor out the constant

$\frac{9}2c\int\limits^3_0 x \, dx = 1$

Integrate

$\frac{9}2c * \frac{x^2}{2}|\limits^3_0 = 1$

Expand

$\frac{9}2c * \frac{3^2}{2} = 1$

$\frac{9}2c * \frac{9}{2} = 1$

Evaluate the product

$\frac{81}{4}c = 1$

Multiply both sides by 4/81

$c = \frac{4}{81}$

Evaluate the quotient

$c = 0.0494$

Hence, the value of c is 0.0494

Read more about probability density function at:

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