Question

) If X and Y each follow a exponential distribution with

parameter 1 and are not independent find the pdf of U=X-Y​

Answer 1

The joint pdf of U = X - Y is $e^{-(x+y)}$   x,y ≥  0

Step-by-step explanation:

Given: X and Y are exponential distribution with parameter 1

To find: The pdf of U=X-Y​

Solution:

X and Y are exponential distribution with parameter 1

We know that ,

If X  ~ Expo( E ) , then the pdf of X is

$F_{x} (X) = \left \{ {{Eex^{-Ex} } \atop {0}} \right.$  x ≥ 0  , otherwise

Here E = 1

The pdf of X is $f_{x} (X) = e^{-x}$ x ≥ 0

The pdf of Y is $f_{y} (Y) = e^{-y}$  y ≥ 0

Since X and Y are independent.

The joint pdf of X and Y is

$f_{xy} (x,y) = f_{x} (x).f_{y} (Y)$

= $e^{-x} .e^{-y}$

= $e^{-(x+y)}$   x,y ≥  0

Final answer:

The joint pdf of U = X - Y is $e^{-(x+y)}$   x,y ≥  0

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Answer 2

Answer:

The pdf of $U=X-Y$ is ​  $e^{-(x+y)}\hspace{1cm} ,\hspace{1cm} x,y\geq 0$

Step-by-step explanation:

Exponential distributions with parameter 1 describe X and Y.

Having said that,

$X$ ~ $Expo( E ),$  so the pdf of $X$ is    

                      $F_{x}(X) =\left \{ {{Eex^{-Ex}} \atop {0}} \right. \hspace{1cm} , \hspace{1cm} x\geq 0$

Since$E = 1$,

PDF of $X$ = $F_{x}(X) ={e^{-x} , \hspace{1cm} x\geq 0$

PDF of $Y$ = $F_{y}(Y) ={e^{-y} , \hspace{1cm} y\geq 0$

Since X and Y are independent,

The joint pdf of X and Y is written as:

$F_{xy}(X,Y) =F_{x}(X).F_{y}(Y)$

                  $=e^{-x} .e^{-y}$

                  $=e^{-(x+y)}\hspace{1cm} ,\hspace{1cm} x,y\geq 0$

                 

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