## Calculating bivariate normal probabilities

This post extends the discussion of the bivariate normal distribution started in this post from a companion blog. Practice problems are given in the next post.

Suppose that the continuous random variables $X$ and $Y$ follow a bivariate normal distribution with parameters $\mu_X$, $\sigma_X$, $\mu_Y$, $\sigma_Y$ and $\rho$. What to make of these five parameters? According to the previous post, we know that

• $\mu_X$ and $\sigma_X$ are the mean and standard deviation of the marginal distribution of $X$,
• $\mu_Y$ and $\sigma_Y$ are the mean and standard deviation of the marginal distribution of $Y$,
• and finally $\rho$ is the correlation coefficient of $X$ and $Y$.

So the five parameters of a bivariate normal distribution are the means and standard deviations of the two marginal distributions and the fifth parameter is the correlation coefficient that serves to connect $X$ and $Y$. If $\rho=0$, then $X$ and $Y$ are simply two independent normal distributions.

When calculating probabilities involving a bivariate normal distribution, keep in mind that both marginal distributions are normal. Furthermore, the conditional distribution of one variable given a value of the other is also normal. Much more can be said about the conditional distributions.

The conditional distribution of $Y$ given $X=x$ is usually denoted by $Y \lvert X=x$ or $Y \lvert x$. In additional to being a normal distribution, it has a mean that is a linear function of $x$ and has a variance that is constant (it does not matter what $x$ is, the variance is always the same). The linear conditional mean and constant variance are given by the following:

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

$\displaystyle Var[Y \lvert X=x]=\sigma_Y^2 \ (1-\rho^2)$

Similarly, the conditional distribution of $X$ given $Y=y$ is usually denoted by $X \lvert Y=y$ or $X \lvert y$. In additional to being a normal distribution, it has a mean that is a linear function of $x$ and has a variance that is constant. The linear conditional mean and constant variance are given by the following:

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

$\displaystyle Var[X \lvert Y=y]=\sigma_X^2 \ (1-\rho^2)$

The information about the conditional distribution of $Y$ on $X=x$ is identical to the information about the conditional distribution of $X$ on $Y=y$, except for the switching of $X$ and $Y$. An example is helpful.

Example 1
Suppose that the continuous random variables $X$ and $Y$ follow a bivariate normal distribution with parameters $\mu_X=10$, $\sigma_X=10$, $\mu_Y=20$, $\sigma_Y=5$ and $\rho=0.6$. The first two parameters are the mean and standard deviation of the marginal distribution of $X$. The next two parameters are the mean and standard deviation of the marginal distribution of $Y$. The parameter $\rho$ is the correlation coefficient of $X$ and $Y$. Both marginal distributions are normal.

Let’s focus on the conditional distribution of $Y$ given $X=x$. It is normally distributed. Its mean and variance are:

\displaystyle \begin{aligned} E[Y \lvert X=x]&=\mu_Y+\rho \ \frac{\sigma_Y}{\sigma_X} \ (x-\mu_X) \\&=20+0.6 \ \frac{5}{10} \ (x-10) \\&=20+0.3 \ (x-10) \\&=17+0.3 \ x \end{aligned}

$\displaystyle \sigma_{Y \lvert x}^2=Var[Y \lvert X=x]=\sigma_Y^2 (1-\rho^2)=25 \ (1-0.6^2)=16$

$\displaystyle \sigma_{Y \lvert x}=4$

The line $y=17+0.3 \ x$ is also called the least squares regression line. It gives the mean of the conditional distribution of $Y$ given $x$. Because $X$ and $Y$ are positively correlated, the least squares line has positive slope. In this case, the larger the $x$, the larger is the mean of $Y$. The standard deviation of $Y$ given $x$ is constant across all possible $x$ values.

With mean and standard deviation known, we can now compute normal probabilities. Suppose the realized value of $X$ is 25. Then the mean of $Y \lvert 25$ is $E[Y \lvert 25]=24.5$. The standard deviation, as indicated above, is 4. In fact, for any other $x$, the standard deviation of $Y \lvert x$ is also 4. Now calculate the probability $P[20. We first calculate it using a normal table found here.

\displaystyle \begin{aligned} P[20

Using a TI84+ calculator, $P[20. In contrast, the probability $P[20 is (using the table found here):

\displaystyle \begin{aligned} P[20

Using a TI84+ calculator, $P[20. Note that $P[20 is for the marginal distribution of $Y$. It is not conditioned on any realized value of $X$.

Practice Problems

Statistics Practice Problems

probability Practice Problems

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## Practice Problem Set 4 – Correlation Coefficient

This post provides practice problems to reinforce the concept of correlation coefficient discussed in this
post in a companion blog. The post in the companion blog shows how to evaluate the covariance $\text{Cov}(X,Y)$ and the correlation coefficient $\rho$ of two continuous random variables $X$ and $Y$. It also discusses the connection between $\rho$ and the regression curve $E[Y \lvert X=x]$ and the least squares regression line.

The structure of the practice problems found here is quite simple. Given a joint density function for a pair of random variables $X$ and $Y$ (with an appropriate region in the xy-plane as support), determine the following four pieces of information.

• The covariance $\text{Cov}(X,Y)$
• The correlation coefficient $\rho$
• The regression curve $E[Y \lvert X=x]$
• The least squares regression line $y=a+b x$

The least squares regression line $y=a+bx$ whose slope $b$ and y-intercept $a$ are given by:

$\displaystyle b=\rho \ \frac{\sigma_Y}{\sigma_X}$

$\displaystyle a=\mu_Y-b \ \mu_X$

where $\mu_X=E[X]$, $\sigma_X^2=Var[X]$, $\mu_Y=E[Y]$ and $\sigma_Y^2=Var[Y]$.

.

For some of the problems, the regression curves $E[Y \lvert X=x]$ coincide with the least squares regression lines. When the regression curve is in a linear form, it coincides with the least squares regression line.

As mentioned, the practice problems are to reinforce the concepts discussed in this post.

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 Practice Problem 4-A $\displaystyle f(x,y)=\frac{3}{4} \ (2-y) \ \ \ \ \ \ \ 0

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 Practice Problem 4-B $\displaystyle f(x,y)=\frac{1}{2} \ \ \ \ \ \ \ \ \ \ \ \ 0

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 Practice Problem 4-C $\displaystyle f(x,y)=\frac{1}{8} \ (x+y) \ \ \ \ \ \ \ \ \ 0

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 Practice Problem 4-D $\displaystyle f(x,y)=\frac{1}{2 \ x^2} \ \ \ \ \ \ \ \ \ \ \ \ \ 0

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 Practice Problem 4-E $\displaystyle f(x,y)=\frac{1}{2} \ (x+y) \ e^{-x-y} \ \ \ \ \ \ \ \ \ \ \ \ \ x>0, \ y>0$

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 Practice Problem 4-F $\displaystyle f(x,y)=\frac{3}{8} \ x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0

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 Practice Problem 4-G $\displaystyle f(x,y)=\frac{1}{2} \ xy \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0

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 Practice Problem 4-H $\displaystyle f(x,y)=\frac{3}{14} \ (xy +x) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0

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 Practice Problem 4-I $\displaystyle f(x,y)=\frac{3}{32} \ (x+y) \ xy \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0

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 Practice Problem 4-J $\displaystyle f(x,y)=\frac{3y}{(x+1)^6} \ \ e^{-y/(x+1)} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x>0, \ y>0$

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 Practice Problem 4-K $\displaystyle f(x,y)=\frac{y}{(x+1)^4} \ \ e^{-y/(x+1)} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x>0, \ y>0$ For this problem, only work on the regression curve $E[Y \lvert X=x]$. Note that $E[X]$ and $Var[X]$ do not exist.

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4-A
• $\displaystyle \text{Cov}(X,Y)=\frac{1}{10}$
• $\displaystyle \rho=\sqrt{\frac{1}{3}}=0.57735$
• $\displaystyle E[Y \lvert X=x]=\frac{2 (4-3 x^2+x^3)}{3 (4- 4x+x^2)}=\frac{2 (2+x-x^2)}{3 (2-x)} \ \ \ \ \ 0
• $\displaystyle y=\frac{2}{3} \ (x+1)$
4-B
• $\displaystyle \text{Cov}(X,Y)=\frac{1}{9}$
• $\displaystyle \rho=\frac{1}{2}$
• $\displaystyle E[Y \lvert X=x]=1+\frac{1}{2} x \ \ \ \ \ 0
• $\displaystyle y=1+\frac{1}{2} x$
4-C
• $\displaystyle \text{Cov}(X,Y)=-\frac{1}{36}$
• $\displaystyle \rho=-\frac{1}{11}$
• $\displaystyle E[Y \lvert X=x]=\frac{x+\frac{4}{3}}{x+1} \ \ \ \ \ 0
• $\displaystyle y=\frac{14}{11}-\frac{1}{11} x$
4-D
• $\displaystyle \text{Cov}(X,Y)=\frac{1}{3}$
• $\displaystyle \rho=\frac{1}{2} \ \sqrt{\frac{15}{7}}=0.7319$
• $\displaystyle E[Y \lvert X=x]=\frac{x^2}{2} \ \ \ \ \ 0
• $\displaystyle y=-\frac{1}{3}+ x$
4-E
• $\displaystyle \text{Cov}(X,Y)=-\frac{1}{4}$
• $\displaystyle \rho=-\frac{1}{7}=-0.1429$
• $\displaystyle E[Y \lvert X=x]=\frac{x+2}{x+1} \ \ \ \ \ x>0$
• $\displaystyle y=\frac{12}{7}-\frac{1}{7} x$
4-F
• $\displaystyle \text{Cov}(X,Y)=-\frac{3}{40}$
• $\displaystyle \rho=\frac{3}{\sqrt{19}}=0.3974$
• $\displaystyle E[Y \lvert X=x]=\frac{x}{2} \ \ \ \ \ 0
• $\displaystyle y=\frac{x}{2}$
4-G
• $\displaystyle \text{Cov}(X,Y)=\frac{16}{225}$
• $\displaystyle \rho=\frac{4}{\sqrt{66}}=0.4924$
• $\displaystyle E[Y \lvert X=x]=\frac{2}{3} x \ \ \ \ \ 0
• $\displaystyle y=\frac{2}{3} x$
4-H
• $\displaystyle \text{Cov}(X,Y)=\frac{298}{3675}$
• $\displaystyle \rho=\frac{149}{3 \sqrt{12259}}=0.4486$
• $\displaystyle E[Y \lvert X=x]=\frac{x (2x+3)}{3x+6} \ \ \ \ \ 0
• $\displaystyle y=-\frac{2}{41}+\frac{149}{246} x$
4-I
• $\displaystyle \text{Cov}(X,Y)=-\frac{1}{144}$
• $\displaystyle \rho=-\frac{5}{139}=-0.03597$
• $\displaystyle E[Y \lvert X=x]=\frac{4x+6}{3x+4} \ \ \ \ \ 0
• $\displaystyle y=\frac{204}{139}-\frac{5}{139} x$
4-J
• $\displaystyle \text{Cov}(X,Y)=\frac{3}{2}$
• $\displaystyle \rho=\frac{1}{\sqrt{3}}=0.57735$
• $\displaystyle E[Y \lvert X=x]=2 (x+1) \ \ \ \ \ x>0$
• $\displaystyle y=2 (x+1)$
4-K
• $\displaystyle E[Y \lvert X=x]=2 (x+1) \ \ \ \ \ x>0$

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