From actuarial exam: Calculate the variance of the retirement package for a new employee, > given that the value is at least 10. 
The distribution of values of the retirement package offered by a
  company to new employees is modeled by the probability density
  function,

Calculate the variance of the retirement package for a new employee,
  given that the value is at least 10.

I know that, Var(Y)= E(V(Y|X>10)+ V(E(Y|X>10)), is an equation that show a relationship between both. But I have no idea how to do this exercise. Thanks, any advise it will be apreciated. 
 A: $$\begin{align*} \operatorname{Var}[X \mid X \ge 10] &= \operatorname{E}[(X - \operatorname{E}[X])^2 \mid X \ge 10] \\ &= \operatorname{E}[X^2 \mid X \ge 10] - \operatorname{E}[X \mid X \ge 10]^2. \end{align*}$$  To this end, we have $$\begin{align*} \operatorname{E}[X^k \mid X \ge 10] &= \frac{1}{\Pr[X \ge 10]} \int_{x=10}^\infty x^k f_X(x) \, dx \\ &= \frac{e}{5} \int_{x=10}^\infty x^k e^{-(x-5)/5} \, dx \\ &= 5^k e \int_{u=1}^\infty (u+1)^k e^{-u} \, du. \end{align*}$$  For $k = 1$, this gives $$\operatorname{E}[X \mid X \ge 10] = 15,$$ and for $k = 2$, this gives $$\operatorname{E}[X^2 \mid X \ge 10] = 250.$$  Therefore, the conditional variance is $250 - 15^2 = 25$.

But an easier way to do this question is to observe that $X$ is a shifted exponential distribution; namely, $X = Y + 5$ where $Y \sim \operatorname{Exponential}(\lambda = 5)$, where $\lambda$ is the rate parameter.  So $$\operatorname{Var}[X \mid X \ge 10] = \operatorname{Var}[Y \mid Y \ge 5] = \operatorname{Var}[Y - 5 \mid Y \ge 5],$$ because the variance is invariant with respect to location.  But an exponential distribution is memoryless, so in particular $$\Pr[Y > y + 5 \mid Y > 5] = \Pr[Y > y],$$ which implies that the variable $$W = Y - 5 \mid Y \ge 5$$ is itself exponential with rate $\lambda = 5$.  Therefore, the variance is simply $\lambda^2 = 25$.
A: Hint: Find the conditional density, then use Double expectation.
