Hi there,

I am not sure if I understood the question correctly. So, take it with a grain of salt and correct me if I am wrong.

From the equation you wrote, It seems to me that given one variable and the sample  you can compute the other variable without any "Random" factor. Therefore, you only have one random variable, the other one is deterministic given the other one. And therefore, you can treat the whole thing as just one random variable.

There is only one probability distribution (for one variable). For the other variable, it is the same probability distribution just with different labels.

I might had missed something here. What kind of equation or expression are you looking for?

Some questions: is the sample an integer? if n > 1, what does Y[1-N] mean?

All the best,

Jhordany

You can google for "tranformation of random variables" and check the results yourself. You will find things like

https://www.cl.cam.ac.uk/teaching/0708/Probabilty/prob11.pdf

or

http://ocw.metu.edu.tr/pluginfile.php/2278/mod_resource/content/0/3.RANDOM_VARIABLES_EXPECTATION_TRANSFORMATION.pdf

Thank you Jhordany for your answer. Sorry for the confusion caused by my lack of details:)

My X and Y variables are the result of a low-pass and high-pass filter respectively. Under the right conditions, these filters are mirrors of each other hence the sum of X & Y at any given time = 1, and their signs are related as shown. This equation is copied from another source, I will edit it to clarify it more.   

Thank you Jochen.

 In what follows I assume that for every  n  there is  a pair of observations ( X[n] , Y[n] ) which satisfy   your equation which is DEPENDENT on  n; as a matter of  fact  you have TWO equations    

X + Y = 1  for  even  n 

- X + Y = 1   for odd  n

Now  if two random variables satisfy the first equation, then their JOINT pd is concentrated on the real line   {(x,y) : x + y  = 1}  \subset R^2.  and consequently  it cannot possess density wrt the two-dimensional Lebesgue measure. However it may have density wrt the linear (one-dimensional) Lebesgue measure on that line.   This happens iff any of  the rv. X  or  Y  possesses density. The relation between the densities, according to the sources linked by  Jochen, is given as follows:

f_X(x)  =  f_Y(1-x)   for  all  x \in  R=(- \infty, \infty).

For presentation of the  joint density , we need to introduce a parameter like the following: the parameter assigned to point    (x , 1-x)  has   value t(x) = x \cdot \sqrt{2}  which shows the distance of the point  on the line from  point  (0,1)  (when we want it to be increasing with  x).  Then the linear joint density equals   

linear-f_{X,Y} (t) =  f_X( t / \sqrt{2} ) /  \sqrt{2} 

(again in accordance to the indicated materials or to any sufficiently reach text-book on probability).

The case of the second equation goes similarly:)

Best regards

Thank you very much Joachim for your answer.

However, can you explain what is meant by cdot in this equation: t(x) = x \cdot \sqrt{2}

Also, would you please elaborate more on this equation: linear-f_{X,Y} (t) = f_X( t / \sqrt{2} ) / \sqrt{2}; I couldn't read it properly.

Thank you.

Dear Hiba,

I have applied the LaTeX convention of encoding mathsymbols:

\cdot  =  the dot which denots multiplication (as of numbers)

\sqrt{m}  = the value of the square root of number m

f_X   =   f with subindex  X, here stands for the density function of the rv X

f_{X,Y} =  f  with double-subindex   X,Y; in my system of notation - the density function of the joint pd. of the pair of random variables  X  and  Y

A \subset  B   =  A is a subset of  B

\infty = the symbol of infinity

x \in A  =  element  x belongs to set  A

\times = cross product ( A \times B denotes usually the cartesian product of two sets  A  and  B, i.e. the set of pairs  (x,y):  x \in A,  y \in  B.

In my answer, symbol  linear-f_{X,Y}   stands for the linear density with respect to parameter   t  (indexing the point on the line  x+y=1). This means for instance, that

the probability that (X,Y)  fails into the rectangle   [1/3; 2]\times [ -4, 1/2]  equals the integral of   linear-f_{X,Y}(t) over  t  running between  indices denoting the ends of the part of the line contained in the rectangle, i.e. between   1/2 \cdot \sqrt{2}  and  2 \cdot \sqrt{2}. Note that it equals the probability, that  X \in [1/2 , 2] = \int f_X(x)  dx between 1/2  and  2. 

Best regards

Dear Joachim,

In your analysis, you said that if the two variables satisfy the first equation and one or both of them have density, then their densities are related by:

 f_X(x) = f_Y(1-x) for all x \in R=(- \infty, \infty)

The same argument goes for the second equation, hence I presume that their densities will be related by:

f_X(x) = f_Y(1+x) for all x \in R=(- \infty, \infty)

If my conclusion is correct, then what does it really mean? in both cases x belongs to R=(- \infty, \infty).

Thank you for your help.

Dear Hiba,

It means, that if  Y= 1 +X  then the graph of the density function  f_Y  is simply shifted by   1   toward positive direction. Everythig happens on the whole real line. Obviously, if   the density of   X   is  zero outside   [0,1] (probably you manage this case)  then the density  f_Y   of the rv  Y  is zero outside   the shifted interval   [0,1]+1 = [1, 2].  For the first case you would have  the density  f_Y   equal zero outside the reflected interval   1 - [0,1] = [0,1].

Hopefully, this explains your problem entirely. For more advanced cases consult the literature on probability, in particular the position indicated above by Jochen

Dear Joachim,

This does clarify the big picture for me. Thank you.