Expected value of stopped iid sequence.

Question

Let $X\ge 0$ a finite expectation continuous random variable with cdf $F$ and denote $\overline{F}=1-F$. Let $\{X_i\}$ a sequence of iid copies of $X$. Let $\tau_p$ be the following stopping rule: stop at $i$ only if $X_i\ge p$. Then: $\mathbb{E} X_{\tau_p}=\mathbb{E} \sum_{i=1}^\infty X_i 1_{\{\tau_p = i\}}=\mathbb{E} X\sum_{i=1}^\infty\mathbb{P}(X_1<p,\ldots,X_{i-1}<p,X_i\ge p)=\mathbb{E} X\overline{F}(p)\sum_{i=1}^\infty F^{i-1}(p)=\mathbb{E} X\frac{\overline{F}(p)}{1-F(p)}=\mathbb{E} X$.

I find the above counterintuitive, since I was expecting to obtain something dependent on $p$. Are there any mistakes in the computation?

EDIT: In the above I forgot to include the last $1_{\{X_i\ge p\}}$ in the expectation, the correct estimate should actually be $\mathbb{E} X_{\tau_p}=\mathbb{E} \sum_{i=1}^\infty X_i 1_{\{\tau_p = i\}}\le \mathbb{E} X\sum_{i=1}^\infty\mathbb{P}(X_1<p,\ldots,X_{i-1}<p)=\mathbb{E} X\sum_{i=1}^\infty F^{i-1}(p)=\frac{\mathbb{E} X}{1-F(p)}$, which does depend on $p$. Is it correct?

Answer 1

The problem is in this step: $$\mathbb{E} \sum_{i=1}^\infty X_i 1_{\{\tau_p = i\}}=\mathbb{E} X\sum_{i=1}^\infty\mathbb{P}(X_1<p,\ldots,X_{i-1}<p,X_i\ge p)$$

You have assumed that the value of $X_i$ in each term can be written as a random variable $X$ that is independent of the event $X_1<p,\dots,X_{i-1}<p,X_i\geq p$. Instead, you should write something like: \begin{align} \mathbb{E} \sum_{i=1}^\infty X_i 1_{\{\tau_p = i\}} &=\sum_{i=1}^\infty\mathbb{E}(1_{\{X_1<p\}}\cdots 1_{\{X_{i-1}<p\}}X_i 1_{\{X_i\ge p\}})\\ &=\sum_{i=1}^\infty F(p)^{i-1}\mathbb{E}(X\, 1_{\{X\ge p\}})\\ &=\mathbb{E}(X\,1_{\{X\ge p\}})\frac{1}{1-F(p)}\\ &=\mathbb{E}(X\mid X\ge p) \end{align} which is what you should expect. The distribution of the first $X_i>p$ is just the conditional distribution of $X$ given $X>p$, and so its expectation is the corresponding conditional expectation.

Answer 2

I use certain definitions of value at risk/tail value at risk in our answer. Different disciplines have different conventions when defining such things (a minus sign vs. no minus sign, inequality flipped, etc.). I will use the definitions I'm familiar with, where the $X_i$ are supposed to represent loss random variables, and adverse events are those when they exceed a threshold. Other disciplines use $X_i$ to model the revenue and look for probabilities that it is low (which is essentially just reflecting and translating). So just be aware that you might find slightly different definitions of the terms I introduce if you search for them.

We let $A\sim B$ mean that $A,B$ have the same distribution.

Since the random variables are continuous, we don't need to distinguish between the events $X_i>p$ and $X_i\geqslant p$.

If $\mathbb{P}(X> p)=0$, then $\tau_p=\infty$ with probability $1$.

If $\mathbb{P}(X> p)=1$, then $\tau_p=1$ with probability $1$.

Assume $p$ is such that $\mathbb{P}(X>p)=\alpha\in(0,1)$. Then $p=\text{VaR}_\alpha(X)$ is the Value at Risk at the $\alpha^{th}$ quantile. Note that $$\mathbb{P}(\tau_p=\infty)=\lim_i \mathbb{P}(\tau_p>i)=\mathbb{P}(X_1,\ldots,X_i\leqslant p)=\lim_i (1-\pi_p)^i=0.$$

It is true that for an event $A$, $\mathbb{E}1_A=\mathbb{P}(A)$. But this does not mean that for a random variable $X$, $\mathbb{E}(X1_A)=\mathbb{E}(X\mathbb{P}(A))$. In fact, since $\mathbb{P}(A)$ is not random, the equality $$\mathbb{E}(X1_A)=\mathbb{E}(X\mathbb{P}(A))=\mathbb{E}(X)\mathbb{P}(A)=\mathbb{E}(X)\mathbb{E}(1_A)$$ holds if and only if $X$ and $A$ are uncorrelated. But $X_i$ and $1_{\{\tau_p=i\}}$ are likely to be correlated. To see an example of this, let $X_i$ be uniform $(-1,1)$, so $\mathbb{E}X=0$. But with $p=0$, the requirement that $\tau_p=i$ requires that $X_i\geqslant 0$. But that's not reflected anywhere in $$\mathbb{E}X\sum_i \mathbb{P}(X_1,\ldots,X_{i-1}\leqslant p,X_i>p).$$ A more extreme example would be $\mathbb{P}(X_i=1)=\mathbb{P}(X_i=-1)=1/2$, where $\mathbb{E}X=0$, but $X_i=1$ whenever $\tau_0=i$. So in this case we do have $$\mathbb{E}X_i1_{\{\tau_0=i\}}=\mathbb{E}1_{\{\tau_0=\}}=\mathbb{P}(X_1,\ldots,X_{i-1}=-1,X_i=1)=2^{-i}.$$ Summing over $i$ gives $\mathbb{E}X_{\tau_0}=1$.

In general, $X_{\tau_p}$ has the same distribution as $X|X>p$. Indeed, fix a measurable subset $B$ of $\mathbb{R}$ and note that \begin{align*} (X_{\tau_p}\in B) & = \bigcup_{i=1}^\infty (X_i\in B,\tau_p=i) = \bigcup_{i=1}^\infty (X_1,\ldots,X_{i-1}\leqslant p)\cap (X_i>p,X_i\in B).\end{align*} The probability of the $i^{th}$ event in this union is $$\mathbb{P}(X\leqslant p)^{i-1}\mathbb{P}(X_i\in B|X_i>p) \mathbb{P}(X_i>p)= (1-\pi)^{i-1}\pi \mathbb{P}(X_i\in B|X_i>p)(1-\pi)^{i-1}\pi \mathbb{P}(X\in B|X>p).$$ Summing over $i$ gives $$\mathbb{P}(X_{\tau_p}\in B)=\mathbb{P}(X\in B|X>p)\sum_{i=1}^\infty (1-\pi)^{i-1}\pi = \mathbb{P}(X\in B|X>p).$$

In particular, $$\mathbb{E}X_{\tau_p}=\mathbb{E}[X|X>p]=\mathbb{E}[X|X>\text{VaR}_\alpha(X)]=\text{TVaR}_\alpha(X),$$ the tail value at risk of $X$ at the $\alpha^{th}$ quantile.