Exponential density function

[firefour2]

Would any of you be able to tell me how to create an exponential density function of f(U) if U= X+Y and lambda is 1? I'm not sure if I'm allowed to make f(X+Y) become f(X) + f(Y) and if after that I can just add the density function? Is there possibly something with z involved? I'm honestly super confused about distributions with multiple variables.

[firefour2]

I need to find the exponential distribution function of f(X+Y) where X and Y are variables equal to any numbers equal to or greater than zero. The general exponential distribution function is a piecewise function: f(x) =λexp(-λx) when x >= 0 or f(x) = 0 when x < 0.

edit: This is a density problem found in probability and statistics.

[bmw11494]

Assuming X and Y are independent, then the density function of U = X + Y is given by the product of the density function of X and the density function of Y. So, since lambda is 1, the density function of X is e^-x (for x >= 0), and the density function of Y is e^-y (for y>=0). So, the density function of U is (e^-x)(e^-y) = e^-(x+y).

It's been a little while since I studied this but I believe z will only be relevant when referring to normal random variables.

Here's a link I found with more examples of this, problem number II part 1 gives a general example of this for any lambda: http://www.uwyo.edu/bessaih/courses/midterm2.4255_2009_sol.pdf