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If $X$ has distribution $\mathbb{exp}(\lambda_1)$ and $Y$ has distribution $\mathbb{exp}(\lambda_2)$, and they are independent of each other, then how do we obtain the MGF of $Z=\mathbb{min}(X,Y)$ directly?

I know that we can obtain the distribution of $Z$ simply by examining its CDF, which tells us that it would be $\mathbb{exp}(\lambda_1 + \lambda_2)$. This appears to be the usual method in finding the minimum's distribution. Easily, this gives us: $$M_Z(t)=\frac{\lambda_1+\lambda_2}{\lambda_1+\lambda_2-t},$$as its MGF. I was curious about doing this by figuring out the expectation instead: $$M_Z(t)=E[e^{t\times\mathbb{min}(X,Y)}].$$ I applied the law of total expectation, partitioning with the events $\{X>Y\}$ and $\{X \le Y\}$. This gives: $$M_Z(t)=E[e^{t\times\mathbb{min}(X,Y)}|X>Y]P(X>Y)+E[e^{t\times\mathbb{min}(X,Y)}|X \le Y]P(X \le Y)$$The conditions on the expected values simplifies the minimums. We also have $ P(X>Y) = \frac{\lambda_2}{\lambda_1+\lambda_2}$. Putting these together, we should get: $$M_Z(t) = E[e^{tY}]\frac{\lambda_2}{\lambda_1+\lambda_2} +E[e^{tX}]\frac{\lambda_1}{\lambda_1+\lambda_2}.$$ Applying the exponential MGF to both, we finally get: $$M_Z(t) = \frac{\lambda_2}{\lambda_2-t} \frac{\lambda_2}{\lambda_1+\lambda_2} + \frac{\lambda_1}{\lambda_1-t}\frac{\lambda_1}{\lambda_1+\lambda_2}.$$To check if this MGF was right, I let $\lambda_1=1$, $\lambda_2=2$ and evaluated the derivative at $t=0$ on Wolfram Alpha. It gave me $2/3$, but I know it's meant to be $1/3$ since it should correspond to the mean: $\frac{1}{\lambda_1+\lambda_2}$. This implies I've messed up somewhere, but I can't find where I've went wrong. Any help would be appreciated.

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You seem to use the identities $E[e^{tY}|X>Y]=E[e^{tY}]$ and $E[e^{tX}|X\leq Y]=E[e^{tX}]$ which are not correct. Instead you can write $$ E[e^{t\min(X,Y)}]=E[e^{tY}1_{\{X>Y\}}]+E[e^{tX}1_{\{X\leq Y\}}]. $$ Now you can calculate each expectation. For example, $$ \begin{align*} E[e^{tY}1_{\{X>Y\}}]&=\int_0^\infty e^{ty}\lambda_2 e^{-\lambda_2 y}\int_y^\infty\lambda_1 e^{-\lambda_1 x}\,\mathrm{d}x\,\mathrm{d}y \\ &=\int_0^\infty e^{ty}\lambda_2 e^{-\lambda_2 y}e^{-\lambda_1 y}\int_0^\infty\lambda_1 e^{-\lambda_1 x}\,\mathrm{d}x\,\mathrm{d}y \\ &=\frac{\lambda_2}{\lambda_1+\lambda_2}\int_0^\infty e^{ty}(\lambda_1+\lambda_2)e^{-(\lambda_1+\lambda_2)}\,\mathrm{d}y \\ &=\frac{\lambda_2}{\lambda_1+\lambda_2}\frac{\lambda_1+\lambda_2}{\lambda_1+\lambda_2-t} \\ &=\frac{\lambda_2}{\lambda_1+\lambda_2-t}. \end{align*} $$ Similarly, $E[e^{tX}1_{\{X\leq Y\}}]=\frac{\lambda_1}{\lambda_1+\lambda_2-t}$. Now you get the desired MGF by adding the two expectations.

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