# Conditional expectation of one random variable

Let $$X$$ be a continuous random variable whose probability density function is $$f(x)= \left\{ \begin{array}{lcc} \alpha^2xe^{-\alpha x} & if & x > 0 \\ \\ 0& otherwise \end{array} \right.$$ With $$\alpha >0$$. Calculate $$E(X|X<1)$$.

I know that by definition, I have $$E(X|X<1)=\int_{0}^{\infty}x \cdot P(X|X<1) dx = \int_{0}^{\infty}x \cdot \frac{P(X=x_k \land X<1 )}{P(X<1)}dx$$ Is ok if i change the limits of the integral and put this? $$E(X|X<1)=\int_{0}^{1}x\cdot \frac{P(X<1 )}{P(X<1)}dx=\frac{1}{2}$$ It confuses me a bit that the condition is on the same random variable. Thanks for the help

• If you're conditioning on an event like $\{X<1\}$ you're essentially using your given pdf $f$ which is originally supported on $(0,\infty)$ to generate a "new" pdf $f_E$ that's supported on $E=(0,1)$. This can be accomplished by considering $f_E(x)=\frac{f(x)}{\int_0^1f(t)dt}$ for $x\in E$ and $f_E(x)=0$ elsewhere. So $$E(X|X<1)=\int_0^1xf_{E}(x)dx$$ – Matthew Pilling 2 days ago

The correct expression is $$\operatorname{E}[X \mid X < 1] = \frac{\int_{x=0}^1 x f_X(x) \, dx}{\int_{x=0}^1 f_X(x) \, dx}.$$ You cannot write $$Pr[X = x_k \wedge X < 1]$$ because first of all, you did not define $$x_k$$, and second, it would need to depend on the variable of integration $$x$$. Then even if you did write $$\Pr[X = x \wedge X < 1]$$, this is problematic because $$X$$ has a density, not a probability mass function. Fourth, even if you ignore all of the above, the events $$X = x$$ and $$X < 1$$ are not independent.