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I have a specific problem I'm working on. Let $X$ be an exponential random variable, and let $Y$ be a random variable defined by:

$$ Y = \begin{cases} 0 & \text{ if } X < d \\ (X - d) & \text{ if } X > d \end{cases} $$

So that $Y$ is $X$ "with a deductible $d$". We are told that $E(Y) = 0.9 E(X)$, and by applying the law of total expectation and memoryless-ness, we can conclude that $P(X>d) = 0.9$.

I've run into a few problems in computing $E(Y^2)$, and I've noticed a possible theorem that would make it all work very neatly. In general, is it true that for a memoryless random variable $X$ and integrable function $g$,

$$ E(g(X - d) \mid X > d) = E(g(X)) ? $$

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$\newcommand{\E}{\mathbb E}$ $$ \E(g(X - d) \mid X > d) = \E(g(X))\text{ ?} $$

For any measurable set $A\subseteq\mathbb R^+$, $$ \Pr(X-d\in A\mid X>d) = \Pr(X\in A). $$ In other words, the conditional probability distribution of $X-d$ given that $X>d$ is the same as the marginal (or "unconditional") probability distribution of $X$.

If two random variables $X$ and $W$ on two different probability spaces have the same distribution, then for any function $g$ for which the expectations are defined, $\E(g(X))=\E(g(W))$.

If $X$ is defined on the space $\Omega$ with measure $P$ then $\{\omega\in\Omega\mid X(\omega)>d\}$ with measure $\dfrac{P}{P(X>d)}$ is a probability space in its own right.

So the answer is affirmative.

Notice that things like $\Pr(X>d)$ and $\E(X)$ depend only on the probabilty distribution of $X$ and not on the underlying space $\Omega$. So if two probability distributions of real random variables are the same, then they have the same expectations, quantiles, etc.

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Show that $f(X=x\mid X>d)=f_X(x)\delta(x>d)/(1-F_X(d))=f_X(x-d)\delta(x>d)$. Then you can write the definition of $\mathbb{E}\left[g(X-d)\mid X>d\right]$ and check whether the two expressions are equal or not.

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  • $\begingroup$ I gave this a try, but I hadn't had exposure to the notation for a density conditional on an arbitrary event (just stuff like $f(y| X = x)$). So I got lost pretty fast. $\endgroup$ – nomen Jul 30 '13 at 0:46

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