# Another Example on Calculating Covariance

In a previous post called An Example on Calculating Covariance, we calculated the covariance and correlation coefficient of a discrete joint distribution where the conditional mean $E(Y \lvert X=x)$ is a linear function of $x$. In this post, we give examples in the continuous case. Problem A is worked out and Problem B is left as exercise.

The examples presented here are also found in the post called Another Example of a Joint Distribution. Some of the needed calculations are found in this previous post.

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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. Calculate the density function, the mean and the variance for the conditional variable $Y \lvert X=x$.
2. Calculate the density function, the mean and the variance for the conditional variable $X \lvert Y=y$.
3. Use the fact that the conditional mean $E(Y \lvert X=x)$ is a linear function of $x$ to calculate the covariance $Cov(X,Y)$ and the correlation coefficient $\rho$.

Problem B
Let $X$ be a random variable with the density function $f_X(x)=4 \ x^3$ where $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. Calculate the density function, the mean and the variance for the conditional variable $Y \lvert X=x$.
2. Calculate the density function, the mean and the variance for the conditional variable $X \lvert Y=y$.
3. Use the fact that the conditional mean $E(Y \lvert X=x)$ is a linear function of $x$ to calculate the covariance $Cov(X,Y)$ and the correlation coefficient $\rho$.

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Background Results

Here’s the idea behind the calculation of correlation coefficient in this post. Suppose $X$ and $Y$ are jointly distributed. When the conditional mean $E(Y \lvert X=x)$ is a linear function of $x$, that is, $E(Y \lvert X=x)=a+bx$ for some constants $a$ and $b$, it can be written as the following:

$\displaystyle E(Y \lvert X=x)=\mu_Y + \rho \ \frac{\sigma_Y}{\sigma_X} \ (x - \mu_X)$

Here, $\mu_X=E(X)$ and $\mu_Y=E(Y)$. The notations $\sigma_X$ and $\sigma_Y$ refer to the standard deviation of $X$ and $Y$, respectively. Of course, $\rho$ refers to the correlation coefficient in the joint distribution of $X$ and $Y$ and is defined by:

$\displaystyle \rho=\frac{Cov(X,Y)}{\sigma_X \ \sigma_Y}$

where $Cov(X,Y)$ is the covariance of $X$ and $Y$ and is defined by

$Cov(X,Y)=E[(X-\mu_X) \ (Y-\mu_Y)]$

or equivalently by $Cov(X,Y)=E(X,Y)-\mu_X \mu_Y$.

Just to make it clear, in the joint distribution of $X$ and $Y$, if the conditional mean $E(X \lvert Y=y)$ is a linear function of $y$, then we have:

$\displaystyle E(X \lvert Y=y)=\mu_X + \rho \ \frac{\sigma_X}{\sigma_Y} \ (y - \mu_Y)$

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Discussion of Problem A

Problem A-1

Since for each $x$, $Y \lvert X=x$ has the uniform distribution $U(0,x)$, we have the following:

$\displaystyle f_{Y \lvert X=x}=\frac{1}{x}$ for $x>0$

$\displaystyle E(Y \lvert X=x)=\frac{x}{2}$

$\displaystyle Var(Y \lvert X=x)=\frac{x^2}{12}$

Problem A-2

In a previous post called Another Example of a Joint Distribution, the joint density function of $X$ and $Y$ is calculated to be: $f_{X,Y}(x,y)=\alpha^2 \ e^{-\alpha x}$. In the same post, the marginal density of $Y$ is calculated to be: $f_Y(y)=\alpha e^{-\alpha y}$ (exponentially distributed). Thus we have:

\displaystyle \begin{aligned} f_{X \lvert Y=y}(x \lvert y)&=\frac{f_{X,Y}(x,y)}{f_Y(y)} \\&=\frac{\alpha^2 \ e^{-\alpha x}}{\alpha \ e^{-\alpha \ y}} \\&=\alpha \ e^{-\alpha \ (x-y)} \text{ where } y

Thus the conditional variable $X \lvert Y=y$ has an exponential distribution that is shifted to the right by the amount $y$. Thus we have:

$\displaystyle E(X \lvert Y=y)=\frac{1}{\alpha}+y$

$\displaystyle Var(Y \lvert X=x)=\frac{1}{\alpha^2}$

Problem A-3

To compute the covariance $Cov(X,Y)$, one approach is to use the definition indicated above (to see this calculation, see Another Example of a Joint Distribution). Here we use the idea that the conditional mean $\displaystyle E(Y \lvert X=x)$ is linear in $x$. From the previous post Another Example of a Joint Distribution, we have:

$\displaystyle \sigma_X=\frac{\sqrt{2}}{\alpha}$

$\displaystyle \sigma_Y=\frac{1}{\alpha}$

Plugging in $\sigma_X$ and $\sigma_Y$, we have the following calculation:

$\displaystyle \rho \ \frac{\sigma_Y}{\sigma_X}=\frac{1}{2}$

$\displaystyle \rho = \frac{\sigma_X}{\sigma_Y} \times \frac{1}{2}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}=0.7071$

$\displaystyle Cov(X,Y)=\rho \ \sigma_X \ \sigma_Y=\frac{1}{\alpha^2}$

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Problem B-1

$\displaystyle E(Y \lvert X=x)=\frac{x}{2}$

$\displaystyle Var(Y \lvert X=x)=\frac{x^2}{12}$

Problem B-2

$\displaystyle f_{X \lvert Y=y}(x \lvert y)=\frac{4 \ x^2}{1-y^3}$ where $0 and $y

Problem B-3

$\displaystyle \rho=\frac{\sqrt{3}}{2 \ \sqrt{7}}=0.3273268$

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

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