If the exponent of $e$ of a bivariate normal density is

$$\frac{-1}{54} *(x^2+4y^2+2xy+2x+8y+4) \\\text{find } \sigma_{1},\sigma_{2} \text{ and } p \text{ given that } \mu_{1} =0 \text{ and } \mu_{2}=-1. $$

One must use this definition to solve.

A pair of random variables $X$ and $Y$ have a bivariate normal distribution and they are referred to as jointly normally distribed random variables if and if their joint probability density is given by

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If $X$ and $Y$ have a bivariate normal distribution normal distribution the conditional density of $Y$ given $X =x$ is a normal distribution with the mean

$\mu_{Y|x} = \mu_{2} + \rho \cdot \frac{\sigma_2}{\sigma_1}\cdot (x-\mu_{1})$

and the variance

$\sigma^{2}_{Y|x} = \sigma^{2}_{2}(1-\rho^2)$

and the conditonal density of $X$ given $Y=y$ is a normal distribution with the mean

$\mu_{X|y} = \mu_{1} + \rho\cdot \frac{\sigma_{1}}{\sigma_{2}}\cdot (y-\mu_{2})$

and the variance

$\sigma^{2}_{X|y} = \sigma^{2}_{1}(1-\rho^2)$

Does one re arrange the formula so that one can just plug in what $\mu_{1} \text{ and } \mu_{2}$ are? Does one substitute like this?

$$f(x,y) = \frac{-1}{54} *(x^2+4y^2+2xy+2x+8y+4) \\ = \frac{1}{-54} \left[ (x-\mu_{1})^2 +2(x-\mu_{1})(y-\mu_{2})+4(y-\mu_{2})^2 \right]$$

EDIT : Thanks to the valiant work of Anatoly I understand one needs to compare coefficients in order to derive $\rho,\sigma_{1},\sigma_{2}$

$$\color{lime}{(1)}: \frac{1}{2(1-p^2)\sigma^2_{1}} = \frac{1}{54}$$

However after this one I do not know what to do does anyone know how to do the rest?


If we substitute $\mu_1=0$ and $\mu_2=-1$ in the exponent of $e $ of the joint probability density formula, we get

$$\displaystyle -\frac 1{2(1-\rho^2)}\left[\left(\frac{x}{\sigma_1}\right)^2 -2\rho \left(\frac{x}{\sigma_1}\right) \left(\frac{y+1}{\sigma_2}\right) +\left(\frac{y+1}{\sigma_2}\right)^2\right]$$

Expanding it, we have

$$-\frac 1{2(1-\rho^2)} \left[\frac{1}{\sigma_1^2}x^2 +\frac{1}{\sigma_2^2} y^2 - \frac{ 2\rho }{\sigma_1 \sigma_2} xy - \frac{ 2\rho }{\sigma_1 \sigma_2} x + \frac{2y}{\sigma_2^2} + \frac{1}{\sigma_2^2} \right]$$

which can be written, if we move a given factor $k^2$ into the brackets, as

$$-\frac {1}{2k^2(1-\rho^2)} \left[\frac{k^2}{\sigma_1^2}x^2 +\frac{k^2}{\sigma_2^2} y^2 - \frac{ 2k^2 \rho }{\sigma_1 \sigma_2} xy - \frac{ 2 k^2\rho }{\sigma_1 \sigma_2} x + \frac{2k^2y}{\sigma_2^2} + \frac{k^2}{\sigma_2^2} \right]$$

Comparing this expression with

$$-\frac {1}{54} \left[x^2 +4 y^2+2 xy +2 x + 8y + 4 \right]$$

which is provided in the OP we directly get $\sigma_1=k \,\,$, $\sigma_2=k/2\,\,$ (note that both variances are by definition positive), and $\rho=-1/2\,\,$. So we have

$$-\frac {1}{2k^2 \left(1-\left (\frac {1}{2} \right)^2 \right)} =-\frac {1}{54} $$

and then $k=\pm 6\,\,$. This leads to $\sigma_1^2=36\,\,$ and $\sigma_2^2=9\,\,$.

  • $\begingroup$ After the "Comparing this expression with" how do you know that $\sigma_{1}=k$ and $\sigma_{2} = \frac{k}{2}$ ? How did you obtain $\rho$? $\endgroup$ – Jon Nov 20 '16 at 0:56
  • 1
    $\begingroup$ You know that the coefficients of $x^2$ and $y^2$ are $1$ and $4$, respectively. This leads to $k^2/\sigma_1^2=1$ and $k^2/\sigma_2^2=4$. Solving these two equations, and knowing that $\sigma_1$ and $ \sigma_2$ are both by definition positive, you get $\sigma_1=k $ and $\sigma_2=k/2$. Then, you know that the coefficient of $xy $ is $2$. So you can write $\displaystyle -\frac {2k^2 \rho}{\sigma_1 \sigma_2}=2$. From this you get $\displaystyle -\frac { 2k^2 \rho}{k^2/2}=2$, which gives $\rho=-1/2$. $\endgroup$ – Anatoly Nov 20 '16 at 9:17

That's pretty much it, though the exponent should be $$ -\frac 1{2(1-\rho^2)}\left[\left(\frac{x-\mu_1}{\sigma_1}\right)^2 -2\rho\left(\frac{x-\mu_1}{\sigma_1}\right)\left(\frac{y-\mu_2}{\sigma_2}\right) +\left(\frac{y-\mu_2}{\sigma_2}\right)^2\right]\tag1 $$ (note the denominator in the factor in front is $(1-\rho^2)$, not $(1-\rho)^2$.)

There's no need to look at conditional distributions, just compare (1) to the given exponent. Employing the common denominator $\sigma_1^2\sigma_2^2$ inside the square brackets, (1) is equal to $$ -\frac 1{2(1-\rho^2)\sigma_1^2\sigma_2^2}\left[\sigma_2^2(x-\mu_1)^2 -2\rho\sigma_1\sigma_2(x-\mu_1)(y-\mu_2) +\sigma_1^2(y-\mu_2)^2\right].\tag2 $$ Now it's a matter of equating coefficients and solving for the unknowns. Notice that the ratio between the coefficient on $x^2$ and the coefficient on $y^2$ is $(\sigma_2/\sigma_1)^2$, so now you have the value for $\sigma_2/\sigma_1$. Also the ratio between the coefficients on $xy$ and $y^2$ is $-2\rho(\sigma_2/\sigma_1)$. This allows you to solve for $\rho$. Knowing the value of $\rho$, you inspect the coefficient on $x^2$ to read off $\sigma_1$, and now you can determine $\sigma_2$.


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