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When I search online for conditional distributions, usually I find tutorials that one random variable is conditioned on another. I want to find some tutorial like the following exercise where there is a random variable that's conditioned on its own value.

X ∼ Exp(1)
Calculate the conditional PDF and CDF of X given X>1
Calculate E[X|X>1] 
Find Var(X|X>1)

What's the mathematical term for this kind of conditional density? Thanks in advance!

Edit: Thanks for the discussions. There are some materials on "Conditioning a Random Variable by an Event" like the following one http://netsys.kaist.ac.kr/lecture/EE210_2017/lecture_notes/ch07_student.pdf. I want to make a summary about the following cases

X discrete
P(X in A1|X in A2)

X continuous
f(x|X in A1)

X discrete, Y discrete
P(X in A|Y in B)

X discrete, Y continuous
P(X in A|Y in B)

X continuous, Y discrete
f(x|Y in B)

X continuous, Y continuous
f(x|y)
Salty Gold Fish
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    The mathematics doesn't really change if you're conditioning on the random variable itself, so it doesn't have a special name as far as I know (just "conditional distribution"). It's the same reason we don't have a different name when the conditioning event is independent, or merely correlated with $X$. – dsaxton Aug 30 '19 at 02:00
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    "$X\gt 1$" is not a random variable: it's an event. It doesn't matter how that event is described or defined. Thus, what you need is a definition of conditioning random variables on events. Presumably you know at least one, for otherwise these tutorials wouldn't even be framing their exercises in these terms. – whuber Aug 30 '19 at 12:43

2 Answers2

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You may be looking at a truncated distribution. E.g.: If $X \sim \mathcal{N}(0,1)$, then a $X$ conditional on $a<X<b$, follows a truncated normal distribution.

https://en.wikipedia.org/wiki/Truncated_normal_distribution

honeybadger
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If $X\sim\mathrm{Expo}(\lambda)$ then by definition of conditional expectation on an event, \begin{align} \mathbb E[X\mid X>1] &= \frac{\mathbb E[X\mathsf 1_{\{X>1\}}]}{\mathbb E[\mathsf 1_{\{X>1\}}]}. \end{align} Since $\mathbb E[\mathsf 1_{\{X>1\}}]=\mathbb P(X>1) = e^{-\lambda}$ and $$ \mathbb E[X\mathsf 1_{\{X>1\}}] = \int_1^\infty \lambda x e^{-\lambda x}\ \mathsf dx = \frac{e^{-\lambda}(1+\lambda)}\lambda , $$ we have $$ \mathbb E[X\mid X>1] = \frac{e^{-\lambda}(1+\lambda)}\lambda\cdot e^\lambda = \frac1\lambda + 1. $$ Alternatively, by the memoryless property of the exponential distribution, we have $$ \mathbb E[X\mid X>1] = \mathbb E[X] + 1 = \frac1\lambda + 1. $$ Similarly, $$ \mathbb E\left[X^2\mathsf 1_{\{X>1\}}\right] = \int_1^\infty \lambda x^2 e^{-\lambda x}\ \mathsf dx = \frac{e^{-\lambda } (\lambda (\lambda +2)+2)}{\lambda ^2}, $$ and so $$ \mathbb E[X^2\mid X>1] = \frac{e^{-\lambda } (\lambda (\lambda +2)+2)}{\lambda^2}\cdot\lambda = \frac{e^{-\lambda } (\lambda (\lambda +2)+2)}{\lambda }. $$ It follows that \begin{align} \operatorname{Var}(X\mid X>1) &= \mathbb E[X^2\mid X>1] - \mathbb E[X\mid X>1]^2\\ &= \frac{e^{-\lambda } (\lambda (\lambda +2)+2)}{\lambda } - \left(\frac1\lambda + 1\right)^2\\ &= \frac{e^{-\lambda } (\lambda +1)^2 (\lambda (\lambda +2)+2)}{\lambda ^3}. \end{align}

Math1000
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