Confused about mathematically (formally) finding a conditional expectation in a double expectation problem versus in words

Question

Textbook example: Losses on an insurance policy are uniformly distributed on [0, 10000]. A policy covering these losses has a 1000 deductible. Calculate the average payment made by this policy for each loss.

Textbook solution: Let $X$ be the loss, $Y = \text{max}\{0, X - 1000\}$ be the payment. Condition on the loss being above or below 1000. Then

$$E[Y] = P[X \leq 1000]E[Y| X \leq 1000] + P[X \geq 1000]E[Y|X \geq 1000].$$

If the loss is below 1000, then the payment is 0, so the mean payment is 0. If the loss is greater than 1000, then the payment is uniform on [0, 9000], so the mean payment is 4500.

$$E[Y] = P[X \leq 1000](0) + P[X \geq 1000](4500)$$

The probability that $X$ is greater than 1000 is $1 - F(1000) = 1 - 1000/10000 = 0.9$. Therefore

$$E[Y] = (0.9)(4500) = 4050.$$

My solution: Now if I was doing this on my own I would have proceeded as follows:

$X$ follows a uniform distribution on [0, 10000] so that the pmf of $X$ is 1/10000. The payment amount $Y$ is a piecewise function defined in terms of $X$. It is $0$ if $X \leq 1000$ and it is $X - 1000$ if $X > 1000$.

Using double expectation we have that

$$E[Y] = E_X[E[Y|X]] = P[X \leq 1000]E[Y| X \leq 1000] + P[X > 1000]E[Y|X > 1000].$$

The probabilities $P[X \leq 1000]$ and $P[X > 1000]$ are gotten by integrating 1/10000 over the intervals $[0,1000]$ and $[1000, 10000]$, respectively, to get $1/10$ and $9/10$.

Question: But now I'm confused about how to formally find the remaining conditional expectations. By my definition of conditional expectation in the continuous case, which is new to me and have never used for inequalities, I would have that

$$E[Y| X \leq 1000] = \int_0^{9000} y \cdot f_{Y|X}(Y|X \leq 1000)\, dy.$$

And for $f_{Y|X}(Y|X \leq 1000)$ I have by my definition of conditional distribution that

$$f_{Y|X}(Y|X \leq 1000) = \frac{f(y,x)}{f_X(x \leq 1000)}.$$

If this is correct then how would I find the numerator? The denominator is just $\int_0^{1000} 1/10000 \, dx = 1/10$.

Answer 1

As you noted, it is much more cumbersome to compute the conditional expectation through the conditional density of $Y$ given $\{X \le 1000\}$ or given $\{X > 1000\}$. It is easier to directly make use of $Y = \max\{0, X-1000\}$ to get $$E[Y \mid X \le 1000] = E[\max\{0, X-1000\} \mid X \le 1000] = 0$$ and $$E[Y \mid X > 1000] = E[\max\{0, X-1000\} \mid X > 1000] = E[X \mid X > 1000] - 1000$$ where you can use the knowledge that the conditional distribution of $X$ given $X > 1000$ is uniform on $[1000, 10000]$. This is essentially the textbook solution.

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Finding the conditional distribution of $Y$ given $\{X \le 1000\}$ or given $\{X > 1000\}$ is a bit thorny because of the mixture of continuous and discrete distributions.

For instance, the conditional distribution of $Y$ given $\{X \le 1000\}$ is actually the non-random quantity $0$, which does not even have a density function, so doing an integral or finding the conditional density does not make sense.

For $Y$ given $\{X > 1000\}$, you can show that the conditional distribution is uniform on $[0, 9000]$, but the way to show this would essentially be using the knowledge that $X$ given $X > 1000$ is uniform on $[1000, 10000]$ and shifting it, which is exactly what the textbook solution does anyway.

Answer 2

If you are going to integrate, don't use Total Expectation unless needed. You don't need it.

$Y$ is a function of $X$, so by the Law of the Unconscious Statastitian:

$$\begin{align}\mathsf E(Y) &=\mathsf E(\max\{0,X-1000\})\\[1ex]&= \int_0^{10000} \max\{0, x-1000\} \cdot\dfrac 1{10000}\,\mathrm d x\\[1ex]&=\int_0^{1000} 0\mathrm d x+\int_{1000}^{10000}\dfrac{x-1000}{10000}\,\mathrm d x\\[1ex]&=4050 \end{align}$$

$\blacksquare$