Hi,

you have to go back to the definition of the joint distribution: F^{YZ}(y,z)=P(Y

If you have PDF of two random variables X and Y and you know that they are dependent and have no further information on that dependence, there is absolutely no way to determine joint PDF of (X,Y). If you could, there would be only one 2D PDF for a pair of 1D distributions.

So you need additional information like the one from the domain you are modeling, sampled data etc.

Of course they are going to be completely correlated. But it is simple to obtain. Do the following calculation:

P(Y;Z)=Average[ delta{Y-g(x)} delta{Z-h(x)} ]

Here the average means OVER P(x)

and delta{...} is the Diract delta function

That it all!

Hi,

if you are interested in the joint density f(y,z), you can marginalize it from the joint f(x,y,z) according to:

f(y,z) = \int f(x,y,z) dx = \int f(y,z|x) * f(x) dx

What you now need is the conditional density f(y,z|x). Here, you can utilize the functional relationship between y,x and z,x according to

f(y,z|x) = \delta(y-g(x)) * \delta(z-h(x))

where \delta is the Dirac delta. Now you can plug in the product of the two Diracs into the above integral and resolve the integral by employing that both h(x) and g(x) are monotonic and revertible functions. There is a theorem which provides closed-form expressions in this cas: see http://mathworld.wolfram.com/DeltaFunction.html equation (7). Plug in the sum and utilize the sifting property of the Dirac. This gives you the solution.

One has to notice that the distribution will be absolutely continuous on the set of Hausdorff dimension one defined as C={y,z; there is x such that y=g(x) and z=h(x)}. Which is a one dimensional curve in two dimension. One can picture a one dimensional density sitting over this line in two dimension. The rest is an integration problem along this line as also mentioned in the other answers.

There is a deterministic relation between Y and Z. A joint probability distribution is not necessary. You can derive a function Y->Z from the given function. At the end is F_y(Y)=F_z(Z) for each realization (Y,Z). This is also the upper bound of a copula function.

I agree with Mathias Raschke. It's not necessary to use a joint probability distribution.

In fact the joint cumulative distribution function $F_{Y,Z}(y,z)$ IS uniquely determined by the hypotheses of your question.

$F_{Y,Z}(y,z)=P(Y \leq y, Z \leq z)=P(g(X) \leq y, h(X) \leq z).$

In case g and h are increasing functions, this becomes

$F_{Y,Z}(y,z)=P(X \leq g^{-1}(y), X \leq h^{-1}(z))=P(X \leq min(g^{-1}(y),h^{-1}(z)))=\int_{-\infty}^{min(g^{-1}(y),h^{-1}(z))} f_X(x)dx$

@Mohammad @Manuel :

It is NOT generally true that if Y=g(X) and Z=h(X) then Y and Z are not independent. For example, take g(x)=constant, then Y is a constant random variable, which is always independent of Z.

It is also interesting to consider the following example regarding a transformation of U and V independent uniform (0,1) random variables. Let

X = cos(2*pi*U)*sqrt(-2*log(V)) = g(U,V)

Y = sin(2*pi*U)*sqrt(-2*log(V)) = h(U,V)

Eventhough X and Y are transformations of the same random vector (U,V) they are independent random variables!

@Mathias @Mohammad :

The Fréchet-Hoeffding upper bound for copulas is also a copula, M(u,v)=min{u,v}, which is the one associated to Z=h(g^(-1)(Y)) as you mentioned, therefore using Sklar's Theorem you may obtain the joint probability distribution of (Y,Z):

F_{YZ}(y,z) = M(F_Y(y),F_Z(z)) = min { F_X(g^(-1)(y)) , F_X(h^(-1)(z)) }

Since M is a singular copula there does not exist a joint density.

Please follow Macro Huber's suggestion. That is the procedure I will use myself.

@Olu:

What for? The question was how to obtain F_YZ and the explicit answer is:

F_YZ(y,z) = min { F_X(g^(-1)(y)) , F_X(h^(-1)(z)) }

Dear Arturo if you look at the Marco Huber comment you would see that HIS commnet is the same answer that I gave before! But in a short way without giving any detail, (going into the P(x,y,z) distribution before takin the marginal distribution, etc.)

The answer is as I said before: P(Y;Z)=Average[ delta{Y-g(x)} delta{Z-h(x)} ]

that is all ! Of course that in order to do the integral you have to take care of many detail concerning g^(-1) and h^(-1) and about the Dirac delta, etc.

In adition your examples are not concerning with this discussion.

1) In your first example if x is a u.d.r.v the result is P(Y,Z)=Delta(g^(-1)[y] - h^(-1)[z])

2) In your second example the Box-Muller transformation concerning two (U,V) r.v.u.d into two (X,Y) e.i.gaussian.r.v, end in this way because of the trigonometric functions.

:-)

FYZ(y,z)=FX(g-1(y)) if g-1(y)h-1(z) from this cdf distribution we can compute the pdf distribution.

fYZ(y,z)=fX(g-1(y)) d(g-1(y))/dy where g-1(y)

Answer from Christophe Ley--the first one--is the best. The joint distribution Y and Z has density function f_X( min ( g^{-1}(y), h^{-1}(z) ) )

The cdf of Y and Z is F_{Y,Z}( y, z ) = Pr ( Y < y and Z < z ) = Pr( g(X)

If g and h are both monotone and invertible, then X, Y and Z are all monotone 1-1 transformations of one another. The joint distribution of two of them is not absolutely continuous (does not admit a joint probability density). But there is also no point in computing the joint probability distribution of, say, Y and Z. It is more efficient to say that Y has such and such a distribution, and Z=h^{-1}( g ( Y ) ) (the composition of the inverse of h with g is another monotone and invertible function).

If you understand the question, you will realize that you don't want the answer.

There are many way to calculate the joint probability distribution. Of course, using the cumulative function as a mathematician like, but there is a simple way to get this joint pdf (of course they a going to be strongly correlated, so you will find a delta function at the end of the calculations [that is way it will not be differentiable]. Do this calculation:

Fyz(y,z)= Average[delta{y-g(x)}  * delta{z-h(x)}]

Where delta=Dirac-delta   and the average[...] means:

Integral [ .... fx(x) dx]

Using that g(x) and h(x) are are monotonically increasing and revertible. You will find a simple expression but with a delta funtion at the end of your calculations.