Conditional probability for an exponential variable Let $X$ be an exponential random variable with parameter $\lambda$. I have to find $\mathbb{E}[X\mid X<\alpha]$ where $\alpha >0$.
I must find $\mathbb{P}(X\mid X<\alpha)$ first.
What I did is as follows:
$\mathbb{P}(X\mid X<\alpha)=\dfrac{\mathbb{P}(X<t \;\text{and}\; X<\alpha)}{\mathbb{P}(X<\alpha)}$, where I assumed $t> \alpha$ then I get the conditional probability to be equal to $1$. I'm not sure if my assumption is right or wrong.
 A: The following is an approach that is more conceptual.
Let the  conditional expectation of $X$ given that $X\lt \alpha$ be $A$. 
Note that the expectation of $X$ given that $X\ge \alpha$ is $\alpha+\frac{1}{\lambda}$. This is a consequence of the fact that by the memorylessness property of the exponential, the additional "waiting time" given that we have waited an amount $\alpha$, is exponentially distributed with parameter $\lambda$. 
It follows by the Law of Total Expectation that 
$$E(X)=\frac{1}{\lambda}=(1-e^{-\lambda\alpha})A+e^{-\lambda\alpha}\left(\alpha+\frac{1}{\lambda}\right).$$
Now we can solve for $A$.
Remark: Your approach from basics also works nicely. Define random variable $Y$ by $\Pr(Y\le y)=0$ if $y\le 0$, $\Pr(Y\le y)=1$ if $y\ge \alpha$, and $\Pr(Y\le y)=\Pr(X\le y|X\lt \alpha)$ otherwise. Then 
$$F_Y(y)=\frac{1-e^{-\lambda y}}{1-e^{-\lambda\alpha}}$$
for $0\lt y\lt \alpha$. Now we can compute $E(Y)$ in several ways. The most basic is to note that $f_Y(y)=\frac{\lambda e^{-\lambda y}}{1-e^{\lambda\alpha}}$ on $(0,\alpha)$ and $0$ elsewhere.  Calculate $\int_0^\alpha y\frac{\lambda e^{-\lambda y}}{1-e^{\lambda\alpha}}\,dy$. 
