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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

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    $\begingroup$ 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 $$E(X|X<1)=\int_0^1xf_{E}(x)dx$$ $\endgroup$ – Matthew Pilling 2 days ago
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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.

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