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The random vector $[\,X \,\,\, Y \,]'$ has probability density function

$f_{X,Y} (x,y) = ke^{-2x^2-3xy-\frac{9}{2}y^2}$, where $k$ is some constant

Find $k.$

Find the marginal probability density functions of $X$ and $Y.$

I know for it to be a valid pdf its integral from negative to positive infinity must be equal to one, and that it must be greater than $0$ for all $x.$ But for starters I'm not sure on the integration.

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  • $\begingroup$ You might need to do some transformation of variables to simplify the integration. See multivariate normal distribution for more information. $\endgroup$
    – Math Lover
    Commented May 28, 2018 at 21:16
  • $\begingroup$ what happened to the question? $\endgroup$
    – Henry
    Commented May 29, 2018 at 14:36

2 Answers 2

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Hint: by completing the square of the exponent, you can integrate the joint density with respect to one variable. For, example

$$f_X(x) = \int_{-\infty}^\infty f_{X,Y}(x,y) \, dy = k \int_{-\infty}^\infty e^{-2 x^2 - 3 xy - 9y^2/2} \, dy = k e^{-3x^2/2} \int_{-\infty}^\infty e^{-\frac{9}{2}(y+ x/3)^2} \, dy = \cdots$$

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A standard form of the bivariate normal density with expected value $(0,0)$ is this: $$ f(x,y) = \frac{1}{2 \pi \sigma_X \sigma_Y \sqrt{1-\rho^2}} \exp\left( -\frac{1}{2(1-\rho^2)}\left[ \frac{x^2}{\sigma_X^2} + \frac{y^2}{\sigma_Y^2} - \frac{2\rho xy}{\sigma_X \sigma_Y} \right] \right) $$ The exponent is this: $$ -\frac 1 {2(1-\rho^2)}\left[ \frac{x^2}{\sigma_X^2} + \frac{y^2}{\sigma_Y^2} - \frac{2\rho xy}{\sigma_X \sigma_Y} \right] = -2x^2-3xy-\frac{9}{2}y^2 $$ So $$ \frac 1 {1-\rho^2} \left[ \left( \frac x {\sigma_X}\right)^2 + \left( \frac y {\sigma_Y} \right)^2 - 2\rho\left( \frac x {\sigma_X} \right)\left( \frac y {\sigma_Y} \right) \right] = 4x^2 + 6xy +9y^2 \tag 1 $$ We can see that this is positive-definite by completing the square and writing it as $$ \left(2x + \frac 3 2 y \right)^2 + \frac {27} 4 y^2, $$ so it's always positive.

Equating coefficients, we get \begin{align} & \frac 1 {(1-\rho^2)\sigma_X^2} = 4 \\[10pt] & \frac 1 {(1-\rho^2)\sigma_Y^2} = 9 \\[10pt] & \frac{-2\rho}{(1-\rho^2)\sigma_X\sigma_Y} = 6 \end{align} From this it follows that \begin{align} \sigma_X^2 & = 1/3, \\[8pt] \sigma_Y^2 & = 4/27, \\[8pt] \sigma_X \sigma_Y & = 2/9, \\[8pt] \rho & = -1/2. \end{align}

The marginal density for $X$ is $$ \frac 1 {\sigma_X\sqrt{2\pi}} \exp\left( \frac {-1} 2 \left( \frac x {\sigma_X} \right)^2 \right) $$ and similarly for $Y.$ (If we had $\mu_X\ne0$ then instead of $\dfrac x {\sigma_X}$ we'd have $\dfrac {x-\mu_X}{\sigma_X}.$)

The constant $k$ should be $\dfrac{1}{2\pi\sigma_X\sigma_Y(1-\rho^2)}.$

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