# PDF of a bivariate normal distribution with correlation coefficient between random variables equals 1

I need help with the line of my thinking, and how to conclude it because I'm unsure about my conclusion. A bivariate normal distribution with correlation coefficient between the random variables=1 does not have a pdf. But suppose we try to sketch the function by finding the probabilities obtained in different intervals where Y=aX+b, for all a,b(which are constants) from the real line (because X and Y are linearly related). If we sketch the function, we get a straight line in 3D over all intervals, we get a probability -say p- over the intervals where X and Y are linearly related (p belongs to [0,1]) and probability everywhere else is 0. My question is, can I conclude that a bivariate normal pdf with correlation coefficient between the random variables=1 can have a function drawn, despite not having a pdf?

• Your ideas about what is meant by a pdf, whether univariate or bivariate, are very much incorrect, and unless you understand what is wrong and correct them, any answers to your question will not make much sense to you. Commented Aug 20, 2016 at 16:19
• Oh right! Probability is measured in intervals for a continuous RV! My bad. Will correct it right away. Thank you. Commented Aug 20, 2016 at 16:21

## 2 Answers

For a bivariate normal PDF, you can visualize the shape in terms of the covariance ellipse. If $x$ and $y$ are uncorrelated and have equal variances, this will be a circle (centered around the 2D mean of the PDF). As you increase the correlation the ellipse will become more anisotropic, in the limit converging to a line segment with orientation +45 degrees and "width" 0. (The principle axes of the ellipse correspond to principle components.)

The resulting infinitely high linear ridge could be considered a PDF or not, depending on the definitions used.

For example, as noted on Wikipedia, the conditional PDF of $X_1$ given that $X_2=x_2$ will be a normal distribution

$X_1\mid X_2=x_2 \ \sim\ \mathcal{N}\left(\mu_1+\frac{\sigma_1}{\sigma_2}\rho( x_2 - \mu_2),\, (1-\rho^2)\sigma_1^2\right)$,

where $\rho$ is the correlation coefficient. So in the perfectly correlated case where $\rho$ goes to 1, the normal distribution has a zero variance. Is this a "PDF"? The limit of a Gaussian as variance goes to zero is in fact one standard way of defining the Dirac delta function. As to whether or not this is a "valid PDF", opinions vary.

Of course in practical terms, the marginal PDFs for $x_1$ and $x_2$ will each just be univariate normal.

• In linear algebra terms, the covariance matrix is rank-deficient, so singular, corresponding to the 0 principle variance (singular value). This means it has a zero determinant, making the normalization factor for the Gaussian infinite (1/0). However the pseudo-inverse of the covariance can be used to define the Mahalanobis distance in exp() argument. So the "likelihood" part could be formally evaluated, I think. Commented Aug 20, 2016 at 17:44
• Thank you very much. That also answered my unasked questions regarding how the correlation affects the graph of a bivariate normal distribution. Commented Aug 20, 2016 at 17:55
• The "resulting infinitely high linear ridge" is an incomplete description of the distribution. You also need to supply a density along that line for it to have any meaning.
– whuber
Commented Aug 21, 2016 at 15:14
• @whuber My comment was as far as I got along those lines mathematically. I think the pseudo-inverse of the covariance matrix should just be $\Sigma^+=\Sigma/\sigma^4$, where $\Sigma$ is the covariance matrix and $\sigma^2$ is the norm of the diagonal of $\Sigma$ (= the non-zero singular value). Commented Aug 21, 2016 at 16:18

Consider two variables defined as $u_1=m+n$ and $u_2=m+b+n$ where $m$ and $b$ are fixed constants and $n$ is normally distributed $n\sim \mathcal{N}(0,\sigma^2)$. Then the joint distribution $p(u_1,u_2|m,b)=p(u_2|u_1,b)p(u_1|m)$ and $u_2$ and $u_1$ are perfectly correlated since $u_2=u_1+b$. We can write the joint distribution then as $p(u_1,u_2|m,b)=\mathcal{N}(u_1;m,\sigma^2)\delta(u_2-(u_1+b))$ where $\delta(\cdot)$ denotes the dirac delta function.