Read this article

http://jaguar.fcav.unesp.br/RME/fasciculos/v28/v28_n1/A8_Carlos.pdf

it might be helpful

The bivariate exponential distribution is not uniquely defined, ie there are several.

Check out

downloads.hindawi.com/journals/mpe/2006/041652.pdf

1- Generating m observations for Uniform U1(0,1) and m observations for Uniform U2(0,1).

2- Generate m observations form exponential distribution by using this transformation

W1j =-((1-rao)/mean1) log(U1j)  ,   j =1,2,...,m, with p.d.f. f1(W1),

f1(W1)= (mean1/(1-rao)) exp(-mean1*W1/(1-rao))    ,     W1 > 0

3- Generate m observations form exponential distribution by using this transformation

W2j =-((1-rao)/mean2) log(U2j)  ,   j =1,2,...,m,   with p.d.f. f2(W2),

f2(W2)= (mean2/(1-rao)) exp(-mean2*W2/(1-rao))    ,     W2 > 0

4- Obtaining one observation of Bivariate exponential (x1,y1) by using:

x1 = sum(w1j) , j=1,2, ..., m

y1 = sum(w2j) , j=1,2, ..., m

5- To obtain n observations (xi,yi) , i=1,2,...n from Bivariate exponential, repeat the previous 4 steps n times.

Best wishes

Huda

Thank you  very  much  Dr. Huda ,I wish all  success in your life.

Best regards

Dear Fadhil,

You can use this valuable paper

Al-Saadi S. D. and    & D.H. Young Department (1980), "Estimators for the correlation coefficient in a bivariate exponential distribution", Journal of Statistical Computation and Simulation. vol.11, pp.13-20.

http://www.tandfonline.com/doi/abs/10.1080/00949658008810386

I wish u all the best

Thank you very much