Let $X1,X2,...Xn$ be iid from $Exponential(1/θ)$. That is $f(x) =\frac{1}{\theta}e^{(\frac{-1}{\theta})} , x > 0 $. Calculate $E[nX(1)]$ where $X(1)$ is the smallest order statistic.
I found the CDF, ($F(X)$), to be $1-e^{(\frac{-x}{\theta})}$
Then: $E[nX(1)]$ = $\\~\\$
$n*E[X(1)]$ = $\\~\\$
$n$*$[ \int_{0}^{\infty} X(1)*n[1-F(X)]^{n-1}*\frac{1}{\theta}e^{(\frac{-1}{\theta})} \,dx]$
Here, I'm using the formula for the pdf of the minimum order statistic: $ n[1-F(X)]^{n-1}*f(x)]$
After making the substitution for $F(X)$ in the integral and doing computation, I am finally left with: $n^2$*$[ \int_{0}^{\infty} X(1)*e^{(x(\frac{n}{\theta}))}\,dx]$
Which I think simplifies to $n^2*(\gamma(2)(\frac{\theta^2}{n^2}))$
Which would then simply be $\theta^2$.
Could someone please verify if my thought process is correct and whether or not there is a simpler way to solve this problem? Thank you.
There is an error in your solution. The density for $X_{(1)}$ is exponential with rate $n/\theta$; i.e., $$f_{X_{(1)}}(x) = n(1 - F_X(x))^{n-1} f_X(x) = n(1-(1-e^{-x/\theta}))^{n-1} \frac{1}{\theta} e^{-x/\theta} = \frac{n}{\theta} e^{-nx/\theta}.$$ You are missing a factor of $1/\theta$ arising from the original density of a single observation $X$.
Then, you have some notational problems: you should instead write
$$\operatorname{E}[n X_{(1)}] = n \int_{x=0}^\infty x f_{X_(1)}(x) \, dx = n \int_{x=0}^\infty x \frac{n}{\theta} e^{-nx/\theta} \, dx = \theta.$$ You should not write $X_{(1)}$ in the integrand.
Rather than performing the integration, it is simpler to note that the minimum order statistic is exponentially distributed with rate $n/\theta$, as stated above; then, its expectation is simply $\theta/n$, hence the expectation of $n X_{(1)}$ is $n (\theta/n) = \theta$.