How to convert a univariate distribution to bivariate distribution

[samrah]

hi
i have MARSHAL-OLKIN exponential weibull distribution which have the following cdf and pdf..
how could i convert it to bivariate distribution?
thanks

 

[FactChecker]

The joint PDF of two independent random variables would just be the product of the two individual PDFs.

 

[samrah]

The joint PDF of two independent random variables would just be the product of the two individual PDFs.

thanks.. i have cdf and pdf (latex codes are given)..how can i make bivariate cdf from this univariate? what is shift parameter in this case? kindly guide me

thanks alot

\begin{align} \label{A5}
F(x) &= \frac{1- e^{-\left(\lambda\, x+\beta\, x^k\right)}}{1-(1-\alpha)\,
e^{-\left(\lambda\, x+\beta\, x^k\right)}}\cdot \boldsymbol I_{(0, \infty)}(x)\,,\\ \label{A6}
f(x) &= \frac{\alpha\,\left(\lambda+ \beta \,k\,x^{k-1}\right)\, e^{-\lambda\,x-\beta\,x^k}}
{\left(1-(1-\alpha )\, e^{-\left(\lambda\, x+\beta\, x^k\right)}\right)^2} \cdot \boldsymbol I_{(0, \infty)}(x)\,,
\qquad \lambda, \beta, k, \alpha > 0 \,;
\end{align}

 

[FactChecker]

You need to decide how you want the two variables to be related. It's easy to determine the joint PDF if they are independent -- just multiply them. Alternatively, you may want to mimic the bivariate normal, where the X and Y variables are not independent, but the vector distance from a center point (mean) is in the same class of distributions of the original distribution. For that, you may want to look at how the two coordinate vectors are related in a bivariate normal (see http://mathworld.wolfram.com/BivariateNormalDistribution.html )

PS. My description of the bivariate normal in terms of a vector "distance" is a loose description, not to be taken literally.

 

[Josh S Thompson]

thanks.. i have cdf and pdf (latex codes are given)..how can i make bivariate cdf from this univariate? what is shift parameter in this case? kindly guide me

thanks alot

\begin{align} \label{A5}
F(x) &= \frac{1- e^{-\left(\lambda\, x+\beta\, x^k\right)}}{1-(1-\alpha)\,
e^{-\left(\lambda\, x+\beta\, x^k\right)}}\cdot \boldsymbol I_{(0, \infty)}(x)\,,\\ \label{A6}
f(x) &= \frac{\alpha\,\left(\lambda+ \beta \,k\,x^{k-1}\right)\, e^{-\lambda\,x-\beta\,x^k}}
{\left(1-(1-\alpha )\, e^{-\left(\lambda\, x+\beta\, x^k\right)}\right)^2} \cdot \boldsymbol I_{(0, \infty)}(x)\,,
\qquad \lambda, \beta, k, \alpha > 0 \,;
\end{align}

what is lambda, Beta, alpha, what is I(0, inf) is k the other variable with x. Is this an advanced question, what's I(0,inf)?

 

[samrah]

what is lambda, Beta, alpha, what is I(0, inf) is k the other variable with x. Is this an advanced question, what's I(0,inf)?

alpha,beta ,lambda and gamma are just parameters... i have to fix three and vary one of them (the location or shift parameter)... i found that alpha is the location parameter...now i have to write these in product form with same this pdf ,beta,lambda and gamma will be fixed and alpha will vary.i don't know about (0,inf)