Pivotal to estimate lambda of a exponential I am studying interval estimation by the method of pivotal quantities. Let $X_1, X_2, ..., X_n$ be a random sample from a p.d.f $f(x;\lambda)=\lambda e^{-\lambda x},  x>0,\lambda >0$.  I have to show that $2\lambda\sum X_i$ is a pivotal for estimating $\lambda$.  
Since $X \sim \exp(\lambda),\ \sum X_i\sim {\rm Gamma}(n,\lambda)$, right? So the distribution of $Q= 2\lambda\sum X_i$ is ${\rm Gamma}(n,\lambda)$. This distribution depends on $\lambda$. Then how can I show that $2\lambda\sum X_i$ is a pivotal for estimating $\lambda$?
 A: 
Since $X~\exp(λ)$, $∑X_i\sim \text{Gamma}(n,λ)$ right?

Yes, that's right (as long as you're using the shape-rate parameterization). 
Let's call that $S$ (i.e. $S=∑X_i$).

so the distribution of $Q=2λ∑X_i$ has $\text{Gamma}(n,λ)$ distribution.

How do you get that? If $S\sim \text{Gamma}(n,λ)$ and $2\lambda S\sim \text{Gamma}(n,λ)\,$, that would imply multiplication by $2\lambda$ had no impact on the distribution.
Does that sound right to you? We know the mean and variance of a random variable are both impacted by multiplication by a constant*, so it doesn't sound right to me.
* $E(kX)=kE(X)$ and $\text{Var}(kX) = k^2 \text{Var}(X)$ by elementary properties of expectation and the definition of variance.
A: One easy way to see the result:
you know that 
$P(X_i > x) = e^{-\lambda x}
$
then$$P(2\lambda X_i > x) = P(X_i > x/2\lambda ) = e^{-x/2}
$$
This proves that the distribution of $Q = 2\lambda\sum X_i$ does not depend on $\lambda$.
Now I wonder how you intend to use this quantity to estimate $\lambda$, as this is not a function of the observations only (you multiply  by $\lambda$!).
