bivariate normal distribution

viraltux

I disagree...

First, two variables can be correlated and still behave normally, so I am not sure what you mean by "as soon as correlation is not zero... resultant marginal distribution is no longer normal."

Second, the example I gave results into a symmetric bivariate distribution, if you accept X|Y=y as normal (and it seems you do) you have to accept Y|X=x as normal too.

learner928

In your example, I believe you are saying the conditional normal distributions having different variance? If that is the case, I don't see how the resultant marginal distribution (from adding all the conditional ones) can still be normal.

Doesn't the conditional distributions need to have the same variance to be able to add up to a normal distribution?

viraltux

Yes, that is what I am saying... The marginal of Y is still normal because I also force the variances of its conditional distributions to change likewise. To understand how the marginal distributions are still normal think about the symmetry of the bivariate distribution.

You could make up more convoluted examples but in this one you only need to see the symmetry, turn the plot 90º and you'll have exactly the same bivariate plot, so you see, if you have no problem accepting X|Y=y as normal, then you should not have any problem accepting Y|X=x as normal too.

learner928

Thanks viraltux, just to confirm my understanding, in your example:

"Imagine that given X=x the variance of the normal conditional distribution of Y is inversely proportional to x, and also imagine that the variance of X condition to Y=y is also inversely proportional to y.
Now you have X,Y, X|Y=y and Y|X=x following normal distributions, but you are getting in the bivariate distribution a contour line that looks nothing like an ellipse which is what you would expect if it would follow a bivariate joint normal distribution... yeah? I let for you the fun to do the formal proof though"

Two things I am not sure about:

1) by setting the variance of the normal conditional distribution of Y inversely proportional to x, is the resultant marginal distribution of Y (by adding up the conditional ones) really still normal despite the changing variance?

2) in your setting, are you sure the joint distribution is no longer bivariate normal, ie. the distribution of Z=X+Y is now not normal?

also since you didn't mention about the mean of the conditional normal distributions, guess your example is only dealing with correlation of zero, I think to create non-zero correlation, we need to shift the mean of the conditional distribution right?

thanks

viraltux

Hi learner,

1) For every Y regardless the value of X, Y is normal, and the same goes for X. So you would be adding up normal distributions all along.

2) Yes, I am sure. For a particular linear combination like X+Y we should check, but I am sure because in a bivariate normal distribution you would expect elliptical contour lines, and the example I gave you has a star-like shape. I made a simulation for you to see the shape and the symmetries I am talking about.

In this example all normal distributions are independent from each other, have μ=0 and their variances change along the axis as described.

Attached Thumbnails

 

learner928

Hi Viraltux,

Can you please tell me within the sample simulation you made, what are the formulae for the change of the variance of the conditional normal distributions?

Thanks a lot!

viraltux

It was something like N(0,1/(0.1+abs(x))) , but if you make it proportional N(0,0.1+abs(x)) you also get non elliptical contour lines.

learner928

If you use that for the change of variance, I think the resultant marginal distribution of Y is no longer normal and would have very high kurtosis (more squeezed around the mean than a normal).

reason is:

If you think in the case of a proper bivariate normal distribution, for each X=x, only by keeping the conditional normal distribution of Y all have the same variance, adding these up produces a marginal distribution of Y which is normal.
Now the only thing we are changing to arrive at your case is to reduce the variance as X goes away from 0 (on both positive and negative side), then naturally the end marginal distribution of Y would be more squashed towards the middle and not be normal anymore?

viraltux

Why would adding Normal distributions would give you anything but another Normal distribution? And why would the kurtosis change at all if for any Normal distribution, regardless its variance, the kurtosis is 0?

learner928

Hi Viraltux,

I think in this case we are not actually doing a sum of normal distributions of independent random variables (which I agree would be normal).

What we are doing is superimposing the conditional normal distributions on each other to get to the marginal distribution of Y.

Let's use an example,

Say X has a probability of 0.5 equal to either x1 or x2.

For X=x1 and X=x2, the conditional distribution of Y is normal with different volatility of say 50% and 10%.

Clearly you can see by superimposing these two conditional distributions of Y, the resultant marginal distribution of Y is clearly not normal as the smaller vol one increases the probability around the mean by more than a normal distribution would.
Marginal distribution of Y would only be normal if the conditional distributions for X=x1 and X=x2 have the same volatility.

hence i think in your case of conditional distributions having different volatilities, the resultant marginal distribution of Y is no longer normal.

Please let me know what you think is wrong above?

Thank you very much again.

viraltux

Ah, OK, I see what you mean, yeah, the conditional distributions would be normal but the marginal would
not. Yet, if you rotate the star 45º you solve these problems, check this:

http://demonstrations.wolfram.com/Ma...iateNormality/

So the bottom line is that normal marginal distributions does not necessarily mean bivariate normal.

Edit: Before you ask, I don't know any counter example when normal conditional distributions and normal marginal distribution occur together... sooo

learner928

cool Viraltux, just so that we are on the same page,

we agree that marginal normal distributions as a condition alone do not have to result in only bivariate normal.

but if we add an extra restriction that all conditional distributions are also normal, you are also not sure whether this will now limit the joint distribution has to be bivariate normal or it can still be something else?

viraltux

Yeah, we are on the same page now, but not sure 100% that these two conditions together means bivariate normal, I feel tempted to say yes though, but unless I see the formal prove showing that these two conditions imply every other condition in the definition of bivariate normal I'd be cautious; for instance, singular matrices are allowed in the definition in which case you would have distributions without density, so there might always be a counter example around the corner.

Anyway if someone has the proof that these two conditions are an equivalent definition of the bivariate normal I want to know too

learner928

Hi Viraltux,

the more I think about it, it does seem the only way of keeping marginal distribution of Y normal is for the normal conditional distributions of Y for each value of X follow the bivariate normal formula, with the mean shifting slightly and keeping variance the same.
As soon as you change the variance of these conditional distributions or shift the mean in a different way, the resultant marginal distribution of Y is no longer normal which would break our condition.

do you think so?

Dickfore

Hey, OP, if the joint probability density function is [itex]\phi(x, y)[/itex], then, how would you express the conditional probability for x, assuming [itex]Y = y_{0}[/itex]?

viraltux

Maybe... or Maybe you can figure out a way to change the variance through the axis in a way that keeps the marginal normal. For instance, increasing the variance is going to create kurtosis in the marginals as we previously discussed, but how about if you increase the variance for a while and then you reduce it back in a way to amount for the increased kurtosis, would you get a normal marginal? So yeah, until I don't see a formal prove I'll keep my healthy 'I don't know' flashing