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Let $X_i$ be exponential with rate $r$, what is the distribution of $\min{\{X_1,X_2-c\}}$ for c>0,<0?

let $g$ be the pdf of $\min{\{X_1,X_2\}}$. is it along the lines of $\frac{g(x) + g(x+c)} {2}$ ?

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    Can you tell us how you came up with $\frac{g(x)+g(x+c)}{2}$? – 2017-02-07
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    so the minimum is $X_1$ 50% of the time and $X_2$ another 50% of the time, hence the 50-50. i am adding a constant $c$ to the pdf, so i displace it to $g(x+c)$ to one of them. doesn't seem correct but seemed promising to me. – 2017-02-07
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    You may want to [edit] your question to put that explanation in the question itself to give a bit more context and what your thoughts are, as that will help people give you a better answer, that is tailored to you. – 2017-02-07

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A r.v. with exponential distribution with rate $\lambda$ has pdf $f(x)=\lambda\exp{(-\lambda x)}$ and cdf $F(x)=1-exp{(-\lambda x)}$.

According to this wiki page on order statistics, if you have $n$ r.v.s with $f$ and $F$, then the minimum has pdf $n(1-F(x))^{n-1}f(x)$.

For $n=2$, this is derived as follows. Let $Y=\min\{X_{1},X_{2}\}$, where $X_{1}$ and $X_{2}$ are i.i.d.. Then $\mathbb{P}[Y\leq y]=\mathbb{P}[\min\{X_{1},X_{2}\}\leq y]=1-\mathbb{P}[\min\{X_{1},X_{2}\}\geq y]=1-(1-F(y))^{2}$.

Finally, when $X$ is exponential then $X-c$ where $c$ is constant just shifts the support. Given this, you can fill in the details.