Distribution function of an exponential random variable I need to determine the distribution function of an exponential r.v. with mean $2,$ given that its outcome is larger than $2$ and give the resulting expected value.
I understand that the pdf is $$f(x, \lambda)=\lambda e^{-\lambda x}; ~x\geq 0.$$
Given that the mean of exponential distribution is $1/\lambda, $ should I say then that $\lambda =1/2?$
I would end with $\lambda e^{-\lambda x} =\frac{1}{2} e^{-0.5x}\geq 2.$ How could I use it ?
 A: Although this is a self-study question, since it is now is over five years old, I'm going to go ahead and give a full answer for expository purposes and to assist later users.
Let's generalise your problem to consider the conditional distribution of $X \sim \text{Exp}(\lambda)$ given $X \geqslant x_0$ for some arbitrary value $x_0 > 0$.  Using the law of conditional probability you have the conditional density:
$$\begin{align}
f(X = x | X \geqslant x_0)
&= \frac{f(X = x, X \geqslant x_0)}{\mathbb{P}(X \geqslant x_0)} \\[6pt]
&= \frac{f(X = x \geqslant x_0)}{\mathbb{P}(X \geqslant x_0)} \\[6pt]
&= \frac{\lambda e^{-\lambda x} \cdot \mathbb{I}(x \geqslant x_0)}{e^{-\lambda x_0}} \\[12pt]
&= \lambda e^{-\lambda (x-x_0)} \cdot \mathbb{I}(x \geqslant x_0) \\[16pt]
&= \text{Exp}(x-x_0|\lambda). \\[6pt]
\end{align}$$
This result tells you that $X | X \geqslant x_0 \sim x_0 + \text{Exp}(\lambda)$, which is called the memoryless property of the exponential distribution.  From this result, you then get the conditional mean:
$$\begin{align}
\mathbb{E}(X | X \geqslant x_0)
&= x_0 + \mathbb{E}(X-x_0 | X \geqslant x_0) \\[6pt]
&= x_0 + \frac{1}{\lambda}. \\[6pt]
\end{align}$$
In the case where $\mathbb{E}(X) = \tfrac{1}{\lambda} = 2$ and $x_0 = 2$ you get $\mathbb{E}(X | X \geqslant x_0) = 2+2=4$, which is again a result of the memoryless property.
A: You're correct in saying that an exponential random variable, $\text{Exp}(\lambda)$, with mean 2 implies $\lambda=\tfrac{1}{2}$.
As has been stated in the comments, the question is asking you to find the probability of $X\leq x$ given $X>2$. It is a conditional probability.
So, you need to find:
$$\text{Pr}(X\leq x\,|\,X>2)$$
This can be achieved using Bayes':
$$\text{Pr}(X\leq x\,|\,X>2)=\frac{\text{Pr}(X\leq x\,,\,X>2)}{\text{Pr}(X>2)}$$
Recall that for an exponential:
$$\text{Pr}(X\leq x)=1-e^{-\lambda x}$$
