PROBABILITY AND STATISTICS FOR ENGINEERING Hossein Sameti Department of Computer Engineering Sharif University of Technology Two Functions of Two Random Variables

Two Functions of Two RV.s  X and Y are two random variables with joint p.d.f  and are functions  define the new random variables:  How to determine the joint p.d.f  with in hand, the marginal p.d.fs and can be easily determined.

Two Functions of Two RV.s  For given z and w,  where is the region in the xy plane such that :

 X and Y are independent uniformly distributed rv.s in  Define  Determine  Obviously both w and z vary in the interval Thus  two cases: Example Solution

 For  With  we obtain  Thus Example - continued

 Also,  and  and are continuous and differentiable functions,  So, it is possible to develop a formula to obtain the joint p.d.f directly. Example - continued

 Let’s consider:  Let us say these are the solutions to the above equations: (a) (b)

Consider the problem of evaluating the probability This can be rewritten as: To translate this probability in terms of we need to evaluate the equivalent region for in the xy plane.

 The point A with coordinates ( z, w ) gets mapped onto the point with coordinates  As z changes to to point B in figure (a), -let represent its image in the xy plane.  As w changes to to C, -let represent its image in the xy plane. (a) (b)

 Finally D goes to  represents the equivalent parallelogram in the XY plane with area  The desired probability can be alternatively expressed as  Equating these, we obtain  To simplify this, we need to evaluate the area of the parallelograms in terms of

 let and denote the inverse transformations, so that  As the point ( z,w ) goes to -the point  Hence x and y of are given by:  Similarly, those of are given by:

 The area of the parallelogram is given by  From the figure and these equations,  so that  or

 This is the Jacobian of the transformation  Using these in, and we get  where represents the Jacobian of the original transformation: ( * )

 Example 9.2: Suppose X and Y are zero mean independent Gaussian r.vs with common variance  Define where  Obtain  Here  Since  if is a solution pair so is Thus Example Solution

 Substituting this into z, we get  and  Thus there are two solution sets   so that Example - continued

 Also is  Notice that here also  Using ( * ),  Thus  which represents a Rayleigh r.v with parameter Example - continued

 Also,  which represents a uniform r.v in  Moreover,  So Z and W are independent.  To summarize, If X and Y are zero mean independent Gaussian random variables with common variance, then - has a Rayleigh distribution - has a uniform distribution. -These two derived r.vs are statistically independent. Example - continued

 Alternatively, with X and Y as independent zero mean Gaussian r.vs with common variance, X + jY represents a complex Gaussian r.v. But  where Z and W are as in except that for the abovementioned, to hold good on the entire complex plane we must have  The magnitude and phase of a complex Gaussian r.v are independent with Rayleigh and uniform distributions respectively. Example - continued

 Let X and Y be independent exponential random variables with common parameter.  Define U = X + Y, V = X - Y.  Find the joint and marginal p.d.f of U and V.  It is given that  Now since u = x + y, v = x - y, always and there is only one solution given by Example Solution

 Moreover the Jacobian of the transformation is given by  and hence represents the joint p.d.f of U and V. This gives  and  Notice that in this case the r.vs U and V are not independent. Example - continued

As we will show, the general transformation formula in ( * ) making use of two functions can be made useful even when only one function is specified.

Auxiliary Variables  Suppose  X and Y : two random variables.  To determine by making use of the above formulation in ( * ), we can define an auxiliary variable  and the p.d.f of Z can be obtained from by proper integration.

 Z = X + Y  Let W = Y so that the transformation is one-to-one and the solution is given by  The Jacobian of the transformation is given by  and hence  or  This reduces to the convolution of and if X and Y are independent random variables. Example

 Let and be independent.  Define  Find the density function of Z.  Making use of the auxiliary variable W = Y, Example Solution

 Using these in ( * ), we obtain  and  Let so that  Notice that as w varies from 0 to 1, u varies from to Example - continued

 Using this in the previous formula, we get  As you can see,  A practical procedure to generate Gaussian random variables is from two independent uniformly distributed random sequences, based on Example - continued

 Let X and Y be independent identically distributed Geometric random variables with  (a) Show that min ( X, Y ) and X – Y are independent random variables.  (b) Show that min ( X, Y ) and max ( X, Y ) – min ( X, Y ) are also independent  (a) Let Z = min ( X, Y ), and W = X – Y.  Note that Z takes only nonnegative values while W takes both positive, zero and negative values Example Solution

 We have P ( Z = m, W = n ) = P { min ( X, Y ) = m, X – Y = n }.  But  Thus Example - continued

represents the joint probability mass function of the random variables Z and W. Also Example - continued

 Thus Z represents a Geometric random variable since and  Note that establishing the independence of the random variables Z and W.  The independence of X – Y and min (X, Y ) when X and Y are  independent Geometric random variables is an interesting observation. Example - continued

 (b) Let Z = min (X, Y ), R = max (X, Y ) – min (X, Y ).  In this case both Z and R take nonnegative integer values  we get Example - continued

 This equation is the joint probability mass function of Z and R.  Also we can obtain:  and  From (9-68)-(9-70), we get  This proves the independence of the random variables Z and R as well. Example - continued