If $X,Y$ $\sim$ $i.i.d.$ $exp(1)$, $X+Y$ $\sim$ $Gamma(2,1)$. I know how to arrive at this using sum of MGFs, but I'm wondering if there's an intuitive way to understand this.
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1Use a convolution. And see the answer to this – Benjamin Wang Jun 18 '22 at 20:51
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1This question is very vague. What kind of "intuitive way" are you looking for? – angryavian Jun 18 '22 at 20:59
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It's hard to be precise, but sometimes you want "more" than a proof. – J.G. Jun 18 '22 at 21:00
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A link between the two distributions is the Poisson process.
Exponential(1) is the time until one arrival in a Poisson process with rate parameter 1.
Gamma(2, 1) is the time until there are two arrivals in a Poisson process with rate parameter 1.
These can be derived by themselves (for example, I ta'd a course where this was showed by using the cdf. Roughly referencing the notes again, let $X$ be a Poisson process with rate 1 so the arrivals in time period of length t is $Exp(t)$. Let T be the time until the 2nd arrival. We have $F(t) = P(T\le t)= 1-P(T>t)=1-P(X_{0,t}=0\text{ or } 1)=1-\sum_{i=0}^1 \frac{e^{-t}t^i}{i!}$. If you take the derivative of this, you are supposed to get the Gamma(2, 1) pdf. You can do the same thing for the time until 1 arrival.)
Because "the time until 2 arrivals" is "the time until one arrival plus the time until one arrival," Gamma(2,1) [time until 2 arrivals] = Exp(1) [time until an arrival] + Exp(1) [time until an arrival]
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