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This question already has an answer here:

If T and S are distributed exponential T,S = Exp(lambda). What is distribution T+S? The solutions says its gamma(2,lambda), but I don't understand why. I can only guess the answer by knowing that E(X+Y) = E(X)+E(Y) and in this case its 1/lambda+1/lambda = 2/lambda. And we know E from gamma distribution that is n/lambda, so T+S is gamma distribution with parameters 2 and lambda. I would also like to ask what in gamma distribution means gamma(nenter image description here)

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marked as duplicate by heropup probability Dec 17 '17 at 8:14

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Proving two distributions are equivalent requires more than proving the first few moments are equal.

We know the MGF (moment generating function) of the Gamma($\alpha$,$\beta$) distribution is:

$$M_X(t)=(1-t/\beta)^{-\alpha}$$

Now take two independent exponential random variables $Y$ and $Z$:

$$Y,Z\sim \text{Exp}(\lambda)$$

Now:

$$\begin{align}M_{Y+Z}(t)&=M_Y(t) M_Z(t)\\ &=\frac{\lambda^2}{(\lambda-t)^2} \end{align}$$

The above expression is the MGF of the Gamma parameterized as follows:

$$Y+Z\sim \text{Gamma}(2,\lambda)$$

What you have circled in red is the Gamma function. It is defined as:

$$\Gamma(z)=\int_{0}^{\infty}x^{z-1}e^{-x}\,dx$$

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