finding $E[X]$ of a exponential distribution with a deductable Losses follow an exponential distribution with parameter $\theta$. For a deductible of 100, the expected payment per loss is 2000. What is the expected payment per loss for a deductible of 500?
They get 
$2,000e^{\frac{-400}{\theta}}$ but I'm not sure how they got this.
I know I need to solve for $\lambda$ so what I did was I transformed the parameter $\theta$ to fit the deductible by subtracting 100. Therefore $Y=\theta-100$. Now to find $E[Y]= \displaystyle\int_{100}^{\infty} \lambda e^{-\lambda \theta +100\lambda}=2000$ but the solution doesn't exist. Where did I go wrong?
Following with what Gaffney said, I integrated $\displaystyle\int_{100}^{\infty}(\theta - 100)\lambda e^{-\theta \lambda}$ which I get $\dfrac{-e^{-100\lambda}}{\lambda}=2000$ (since the $-\theta e^{-\theta \lambda} + 100e^{-\theta \lambda}$ will cancel out with 100) substituting 500 gives us $-400e^{-500\lambda} - \dfrac{e^{-500\lambda}}{\lambda}= ?$ which we are trying to figure out but I do not know how to make the connection of the 1st result with the 2nd.

If the loss $X$ is $\lt 100$, then the payment is $0$. If $X\ge 100$, the payment is $X-100$. Thus the expected payment is
$$\int_{100}^\infty (x-100)\frac{1}{\theta} e^{-x/\theta}\,dx.$$
This turns out to be $\theta e^{-100/\theta}$.
Similarly, the expected value of the payment if the deductible is $500$ turns out to be $\theta e^{-500/\theta}$.
Putting $\theta e^{-100/\theta}=2000$ we get the expression given as an answer.
A more complete answer can be given by solving $\theta e^{-100/\theta}=2000$ numerically. We get that $\theta$ is approximately $2097.65$. From this we can find the numerical value of the expected payment with a deductible of $500$. 
