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I need to calculate the distribution of X and Y knowing that X is uniformly distributed in [0,2] and Y is exponentially distributed with variance $1/4$, when X and Y are independent.

So first I've inferred the mean of Y knowing that $Var(Y) = 1/λ^2 = 1/4$ so $λ = 2$ and $E(Y)= 1/2$

I know that $E(X) = (a+b)/2 = 1 $ and $Var(X) = (b-a)^2/12 = 1/3$

I know that $E(X + Y) = E(X) + E(Y)$ and $Var(X+Y) = Var(X) + Var(Y) + 2*Cov(X,Y) = Var(X) + Var(Y)$ because of the independence of X and Y.

Should I simply answer that:

  • $E(X+Y) = 1 + 1/2 = 3/2 $ and $Var(X+Y) = 1/3 + 1/2 = 5/6$ ?

In another example I've found they used double integrals and minimum so I'm not sure anymore if my method is right

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    Your method is correct, although $\mathrm{E}(X+Y) = 1/3 + 1/2$ (you made a typo) – 2017-01-10
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    E(X) ist not equal to $(a+b)/2 = (2+0)/2 = 1$ ? – 2017-01-10
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    Correct, I made a typo myself. So $\mathrm{E}(X+Y) = 1 + 1/2$, which is what I meant to comment. – 2017-01-10
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    Right ! I don't know why I made 1*(1/2) in my head. Thanks :) – 2017-01-10
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    Your 'method' does not answer the question at all. What has finding the mean or variance of the sum got to do with finding the distribution of the sum? – 2017-01-10

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