We refer again to the pair of continuous variables X,Y of fx,y(x,y)=l2exp⁡[−lx]for0&lt;y&lt;x&lt;∞forsome parameter l&gt;0. Consider the transformation U = X – Y and V = Y. 1.Determine the joint pdf of U an V using the Jacobian of the transformation, the support of U and V, etc. Do not forget the support. 1.Are U and V independent? What are their marginal probability density functions and parameters? They are gamma (U) and exponential (V)

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We refer again to the pair of continuous variables X,Y of fx,y(x,y)=l2exp⁡[−lx]for0&amp;lt;y&amp;lt;x&amp;lt;∞forsome parameter l&amp;gt;0. Consider the transformation U = X – Y and V = Y. 1.Determine the joint pdf of U an V using the Jacobian of the transformation, the support of U and V, etc. Do not forget the support. 1.Are U and V independent? What are their marginal probability density functions and parameters? They are gamma (U) and exponential (V)

Jozlyn · Accepted Answer

Step 1 Let us consider the transformation U=X-Y...(1) &amp;amp; V=Y...(2) 1) Using the Jacobin of the transformation, we have to determine the joint pdf of U and V. From, (1) and (2) X=U+V &amp;amp; Y=V Now, J(x,yu,v)=1101=1 FU,V(u,v)=FX,Y(x,y) ={λ2e−λ(u+v),for0&amp;lt;v&amp;lt;u&amp;lt;v&amp;lt;∞0,otherwise Step 2 2) FU(u)=∫−∞∞FU,V(u,v)dv =∫0∞λ2e−λue−λvdv =λe−λu[−e−λv]{0}∞ =λe−λu;0&amp;lt;u&amp;lt;∞ FV(u)=∫−∞∞FU,V(u,v)du =∫0∞λ2e−λue−λvdu =λe−λv[−e−λu]{0}∞ =λe−λv;0&amp;lt;v&amp;lt;∞ ∵FU(u)FV(u)=FU,V(u,v)∀u,v Hence, U&amp;amp;V are independent

We refer again to the pair of continuous variables X,Y of f_{x,y}(x,y

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