Bivariate Normal Distribution Mean

$X$ and $Y$ have bivariate normal distribution and have joint pdf
\begin{equation*} f\left( x,y\right) =a\exp \left( \frac{-1}{2}\omega \right) ,\text{where }% \omega =6x^{2}+12y^{2}-16xy-8x+24 \end{equation*} then what are the means of $X$ and $Y$?

I have \begin{equation*} \frac{1}{\left( 1-\rho ^{2}\right) \sigma _{X}^{2}}=6,~\frac{1}{\left( 1-\rho ^{2}\right) \sigma _{Y}^{2}}=12~\text{and }\frac{-2\rho }{\left( 1-\rho ^{2}\right) \sigma _{X}\sigma _{Y}}=-16 \end{equation*} so i can find $\rho ,\sigma _{X}$ and $\sigma _{Y}$ using these equations. But how can i get $\mu _{X}$ and $\mu _{Y}$?

this question asked in an exam and requested to be replied in 2 minutes. Is there a shortcut for this?

• Match the theoretical expression of a joint density of two correlated normals, to what you have, and see where it gets you. – Alecos Papadopoulos Apr 15 '14 at 14:09
• i have already tried that way and i know i can get the result using matching. But this question asked in an exam and requested to be replied in 2 minutes. – mert Apr 15 '14 at 14:16
• Do the means coincide with the point where the density is maximized? If so you have a shortcut – Adrian Apr 15 '14 at 16:18
• Just minimize the function $\omega$ over (x, y), which I think gives you (6, 4) – Adrian Apr 15 '14 at 16:19
• I think this works: en.wikipedia.org/wiki/Multivariate_normal_distribution the mode of a multivariate normal is $\mu$ – Adrian Apr 15 '14 at 16:25

2 Answers

The means coincide with the point where the density is maximized. See en.wikipedia.org/wiki/Multivariate_normal_distribution: the mode of a multivariate normal is $\mu$.

The shortcut is therefore to minimize the function ω over $(x, y)$, which gives $\mu = (6, 4)$.

Hint: As Adrian points out, we need to minimize $\omega$ as a function of $x$ and $y$. Now, \begin{align} \omega(x) &= 6x^2 - x(16y+8) + \cdots\\ \omega(y) &= 12y^2 - y(16x) + \cdots \end{align} have minima at $$x = \frac{16y+8}{12} = \frac 43y + \frac 23, \quad y = \frac{16x}{24} = \frac{2}{3}x$$ respectively, giving $(x,y) = (6,4)$ as the solution. Here, instead of the calculus-based method of finding minima, I used the fact that $ax^2+bx+c$ has an extremum at the average value of its roots $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ which average equals $\frac{-b}{2a}$ (no need to fuss with evaluating the square root). I leave it up to you to determine if this is doable in 2 minutes or not.