Exponential Random Variables and either cases of a Conditional Expectation We are given a random variable X which has an exponential distribution of parameter λ=1.
$$X\sim\exp(λ=1)$$
We know that $$E[X]=\frac{1}{λ}$$ 
Hence for us $E[X]=1$. By virtue of the memoryless property of exponential distribution, I have understood that:
$$E[X\mid X>x]=x+E[X]=x+\frac{1}{λ}=x+1$$
but what about the either case:
$$E[X\mid X\le x]= ??$$ 
I suppose something like:
$$E[X\mid X\le x]=x-E[X]=x-1$$
but then if $x$ is smaller than $E[X]$ (in this case $E[X]=1$) it would result in a negative expectation. Is that correct or such either case is not expressed like I did?
 A: We have 
$$E(X)=E(X|X\le k)\Pr(X\le k)+E(X|X\gt k)\Pr(X\gt k).\tag{1}$$
You know $E(X)$, $E(X|X\gt k)$ and the probabilities mentioned in (1). Now $E(X|X\le k)$ is determined.
There are other ways to solve the problem, for example by finding the conditional distribution of $X$ given $X\le k$.
A: To check if your intuition is correct, you can use that the density of $X$ given that $X\in A$ for a set $A$ with $P(X\in A)>0$ is equal to $$f_{X\mid X\in A}(u\mid X\in A)=\begin{cases}\frac{f_x(u)}{P(X\in A)}, & u\in A\\ 0, & u \notin A\end{cases}$$ So in your case $$f_{X\mid X\le x}(u\mid X\le x)=\frac{f_X(u)}{P(X\le x)}=\frac{e^{-u}}{1-e^{-x}}$$ for $u\le x$. Thus $$\begin{align*}E[X|X\le x]&=\int_{0}^{x}uf_{X|X\le x}(u\mid X\le x) du =\int_{0}^{c}u\frac{e^{-u}}{1-e^{-x}}du=\frac{1}{1-e^{-x}}\int_{0}^{x}ue^{-u}du=\\\\&=\frac{1-e^{-x}(x+1)}{1-e^{-x}}=1-\frac{x}{e^{x}-1}\end{align*}$$ 
A: An indirect approach follows from the total expectation formula, since you know $\mathbb{E}(X)$, $\mathbb{P}(X > x)$, and $\mathbb{E}(X > x)$ already. A direct approach works too:
$$\mathbb{E}(X|X \leq x) = \frac{\int_0^x y e^{-y} dy}{\int_0^x e^{-y} dy}.$$
In general we have
$$\mathbb{E}(X | A) = \frac{\mathbb{E}(X 1_A)}{\mathbb{P}(A)}$$
for any random variable $X$ and any event $A$ such that $\mathbb{P}(A) > 0$. When $X=1_B$ this is the usual definition of conditional probability.
