Tag Archives: Uniform Distribution

Another Example of a Joint Distribution

In an earlier post called An Example of a Joint Distribution, we worked a problem involving a joint distribution that is constructed from taking product of a conditional distribution and a marginial distribution (both discrete distributions). In this post, we work on similar problems for the continuous case. We work problem A. Problem B is left as exercises.

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Problem A
Let X be a random variable with the density function f_X(x)=\alpha^2 \ x \ e^{-\alpha x} where x>0. For each realized value X=x, the conditional variable Y \lvert X=x is uniformly distributed over the interval (0,x), denoted symbolically by Y \lvert X=x \sim U(0,x). Obtain solutions for the following:

  1. Discuss the joint density function for X and Y.
  2. Calculate the marginal distribution of X, in particular the mean and variance.
  3. Calculate the marginal distribution of Y, in particular, the density function, mean and variance.
  4. Use the joint density in part A-1 to calculate the covariance Cov(X,Y) and the correlation coefficient \rho.

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Problem B
Let X be a random variable with the density function f_X(x)=4 \ x^3 where 0<x<1. For each realized value X=x, the conditional variable Y \lvert X=x is uniformly distributed over the interval (0,x), denoted symbolically by Y \lvert X=x \sim U(0,x). Obtain solutions for the following:

  1. Discuss the joint density function for X and Y.
  2. Calculate the marginal distribution of X, in particular the mean and variance.
  3. Calculate the marginal distribution of Y, in particular, the density function, mean and variance.
  4. Use the joint density in part B-1 to calculate the covariance Cov(X,Y) and the correlation coefficient \rho.

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Discussion of Problem A

Problem A-1

The support of the joint density function f_{X,Y}(x,y) is the unbounded lower triangle in the xy-plane (see the shaded region in green in the figure below).

Figure 1

The unbounded green region consists of vertical lines: for each x>0, y ranges from 0 to x (the red vertical line in the figure below is one such line).

Figure 2

For each point (x,y) in each vertical line, we assign a density value f_{X,Y}(x,y) which is a positive number. Taken together these density values sum to 1.0 and describe the behavior of the variables X and Y across the green region. If a realized value of X is x, then the conditional density function of Y \lvert X=x is:

    \displaystyle f_{Y \lvert X=x}(y \lvert x)=\frac{f_{X,Y}(x,y)}{f_X(x)}

Thus we have f_{X,Y}(x,y) = f_{Y \lvert X=x}(y \lvert x) \times f_X(x). In our problem at hand, the joint density function is:

    \displaystyle \begin{aligned} f_{X,Y}(x,y)&=f_{Y \lvert X=x}(y \lvert x) \times f_X(x) \\&=\frac{1}{x} \times \alpha^2 \ x \ e^{-\alpha x} \\&=\alpha^2 \ e^{-\alpha x}  \end{aligned}

As indicated above, the support of f_{X,Y}(x,y) is the region x>0 and 0<y<x (the region shaded green in the above figures).

Problem A-2

The unconditional density function of X is f_X(x)=\alpha^2 \ x \ e^{-\alpha x} (given above in the problem) is the density function of the sum of two independent exponential variables with the common density f(x)=\alpha e^{-\alpha x} (see this blog post for the derivation using convolution method). Since X is the independent sum of two identical exponential distributions, the mean and variance of X is twice that of the same item of the exponential distribution. We have:

    \displaystyle E(X)=\frac{2}{\alpha}

    \displaystyle Var(X)=\frac{2}{\alpha^2}

Problem A-3

To find the marginal density of Y, for each applicable y, we need to sum out the x. According to the following figure, for each y, we sum out all x values in a horizontal line such that y<x<\infty (see the blue horizontal line).

Figure 3

Thus we have:

    \displaystyle \begin{aligned} f_Y(y)&=\int_y^\infty f_{X,Y}(x,y) \ dy \ dx \\&=\int_y^\infty \alpha^2 \ e^{-\alpha x} \ dy \ dx \\&=\alpha \int_y^\infty \alpha \ e^{-\alpha x} \ dy \ dx \\&= \alpha e^{-\alpha y}  \end{aligned}

Thus the marginal distribution of Y is an exponential distribution. The mean and variance of Y are:

    \displaystyle E(Y)=\frac{1}{\alpha}

    \displaystyle Var(Y)=\frac{1}{\alpha^2}

Problem A-4

The covariance of X and Y is defined as Cov(X,Y)=E[(X-\mu_X) (Y-\mu_Y)], which is equivalent to:

    \displaystyle Cov(X,Y)=E(X Y)-\mu_X \mu_Y

where \mu_X=E(X) and \mu_Y=E(Y). Knowing the joint density f_{X,Y}(x,y), we can calculate Cov(X,Y) directly. We have:

    \displaystyle \begin{aligned} E(X Y)&=\int_0^\infty \int_0^x  xy \ f_{X,Y}(x,y) \ dy \ dx \\&=\int_0^\infty \int_0^x xy \ \alpha^2 \ e^{-\alpha x} \ dy \ dx \\&=\int_0^\infty \frac{\alpha^2}{2} \ x^3 \ e^{-\alpha x} \ dy \ dx \\&= \frac{3}{\alpha^2} \int_0^\infty \frac{\alpha^4}{3!} \ x^{4-1} \ e^{-\alpha x} \ dy \ dx \\&= \frac{3}{\alpha^2} \end{aligned}

Note that the last integrand in the last integral in the above derivation is that of a Gamma distribution (hence the integral is 1.0). Now the covariance of X and Y is:

    \displaystyle Cov(X,Y)=\frac{3}{\alpha^2}-\frac{2}{\alpha} \frac{1}{\alpha}=\frac{1}{\alpha^2}

The following is the calculation of the correlation coefficient:

    \displaystyle \begin{aligned} \rho&=\frac{Cov(X,Y)}{\sigma_X \ \sigma_Y} = \frac{\displaystyle \frac{1}{\alpha^2}}{\displaystyle \frac{\sqrt{2}}{\alpha} \ \frac{1}{\alpha}} \\&=\frac{1}{\sqrt{2}} = 0.7071 \end{aligned}

Even without the calculation of \rho, we know that X and Y are positively and quite strongly correlated. The conditional distribution of Y \lvert X=x is U(0,x) which increases with x. The calculation of Cov(X,Y) and \rho confirms our observation.

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Answers for Problem B

Problem B-1

    \displaystyle f_{X,Y}(x,y)=4 \ x^2 where x>0, and 0<y<x.

Problem B-2

    \displaystyle E(X)=\frac{4}{5}
    \displaystyle Var(X)=\frac{2}{75}

Problem B-3

    \displaystyle f_Y(y)=\frac{4}{3} \ (1- y^3)

    \displaystyle E(Y)=\frac{2}{5}

    \displaystyle Var(Y)=\frac{14}{225}

Problem B-4

    \displaystyle Cov(X,Y)=\frac{1}{75}

    \displaystyle \rho = \frac{\sqrt{3}}{2 \sqrt{7}}=0.327327

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