What is $E(X\mid X>c)$ in terms of $P(X>c)$?
I've seen conditional probability/expectation before with respect to another random variable but not to the variable itself. How would I go about interpreting this?
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4$\begingroup$ As a simple example, suppose $X$ is a discrete random variable taking on values $0$, $1$, $2$ with equal probability $\frac{1}{3}$. Then, conditioned on $X > 0$, $X$ can take on values $1$ or $2$ with probability $$P\{X = i\mid X > 0\} = \frac{P(\{X = i\}\cap \{X > 0\})}{P\{X > 0\}} = \frac{P\{X = i\}}{P\{X > 0\}} = \frac{1/3}{2/3} = \frac{1}{2}, i = 1, 2$$ and so $$E[X\mid X > 0]= \frac{1}{2}\times[1 + 2] = \frac{3}{2}$$. $\endgroup$– Dilip SarwateCommented Dec 8, 2011 at 2:22
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2 Answers
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$$ \mathrm E(X\mid X\gt c)=\frac{\mathrm E(X\cdot\mathbf 1_{X\gt c})}{\mathrm P(X\gt c)} $$
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1$\begingroup$ Could you please clarify what the conditional density function and conditional cumulative distribution would be for this case? $\endgroup$– texmexCommented Mar 14, 2016 at 3:09
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1$\begingroup$ @user249613 If you have a new problem, you might want to post it as a new question. If you do so, adding your thoughts on the subject would be mandatory. $\endgroup$– DidCommented Mar 14, 2016 at 6:10
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@Dilip has already provided an example of how to compute $E(X|X>c)$ in the discrete case. In the continuous case, the computation will be:
$$E(X|X>c) = \frac{\int_x x f(x) I(x>c) dx}{\int_x f(x) I(x>c) dx}$$
Thus, if we let $X \sim U[0,1]$ and $c=0.5$ then we have:
$$E(X|X>0.5) = \frac{\int_x x I(x>0.5) dx}{\int_x I(x>0.5) dx} = \frac{(1 - 0.5^2)/2}{0.5}=0.75$$
In terms of interpretation: Intuitively, you can think of $E(X|X>c)$ as the average of all $X$ values subject to each one of them being greater than $c$.
