Changing expectations into probability.

Question

Going through a proof (related to probability) I found on a paper(Comp sci.) i could not undertstand the steps he followed. Basically the author did the following steps:

$$E(T_{Fail}) = E(t_{f_{\scriptstyle i}} - t_{k} | t_{k} < t_{f_{\scriptstyle i}} \le t_{k + 1})$$

$$E(T_{Fail}) = \int^{t_{k+1} - t_{k}}_{0} P(\tau > t_{f_{\scriptstyle i}} - {t}_{k} | t_{k} < t_{f_{\scriptstyle i}} \le t_{k + 1}) d \tau$$

$\tau$ is not defined anywhere throughout the paper.

There are more steps after the above-mentioned lines(too much to type).

So i don't know how we are moving from expectation to probability. The link to the paper is: http://nslab.kaist.ac.kr/courses/2014/cs712/paperlist/2-17.pdf

Page:19 and 20. Thank you.

Answer 1

$\tau$ is not defined anywhere throughout the paper.

Not true. It's defined right here:

$$\int_0^{t_{k+1} - t_k} P(\tau > t_{f_{\scriptstyle i}} - t_k \mid t_k < t_{f_{\scriptstyle i}} \le t_{k + 1})\, d \tau. \tag1$$

The notation $d\tau$ inside an integral like this tells you that $\tau$ is the variable over which the integral is integrated. That's all it is, and this definition applies only inside the integral.

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I believe there is an error in the paper. It is possible that it is canceled out by another error, but the following is as far as I have gotten at this time:

It is a fact (which the authors of the paper assume the reader will know) that if $X$ is a non-negative random variable with a finite expectation, then $$ E(X) = \int_0^\infty P(X>\tau)\, d\tau. \tag2$$ Refer to the answers to Explain why $E(X) = \int_0^\infty (1-F_X (t)) \, dt$ for every nonnegative random variable $X$ to see why.

In the paper, you have the expectation of the random quantity $t_{f_{\scriptstyle i}} - t_k$ conditioned on $t_k < t_{f_{\scriptstyle i}} \le t_{k + 1},$ where $t_k$ and $t_{k+1}$ are non-random numbers. Since the condition implies that $t_{f_{\scriptstyle i}} - t_k > 0,$ the conditional distribution is that of a non-negative variable, so we can use Equation $(2),$ but we have to write it in terms of the condition $t_k < t_{f_{\scriptstyle i}} \le t_{k + 1}$:

$$ E(t_{f_{\scriptstyle i}} - t_k \mid t_k < t_{f_{\scriptstyle i}} \le t_{k + 1}) = \int_0^\infty P(t_{f_{\scriptstyle i}} - t_k > \tau \mid t_k < t_{f_{\scriptstyle i}} \le t_{k + 1})\, d\tau. \tag3$$

Now we just need to notice that $P(t_{f_{\scriptstyle i}} - t_k > \tau \mid t_k < t_{f_{\scriptstyle i}} \le t_{k + 1}) = 0$ whenever $\tau > t_{k+1} - t_k,$ so we can change the upper end of the integral from $\infty$ to $t_{k+1} - t_k$ without changing the value of the integeral. and that $t_{f_{\scriptstyle i}} - t_k > \tau$ is equivalent to $\tau < t_{f_{\scriptstyle i}} - t_k.$ Therefore the integral on the right-hand side of Equation $(3)$ would be equal to the integral in Expression $(1),$ except for one thing: in Expression $(1),$ we see $\tau < t_{f_{\scriptstyle i}} - t_k,$ which (as you observed) is not equivalent to $t_{f_{\scriptstyle i}} - t_k > \tau.$

It seems easy enough to show that the formula in the paper is wrong. A single counterexample will suffice. Let $t_k = 0,$ $t_{k+1} = 1,$ and suppose the failure (if there is one in that interval) invariably occurs in the interval $(0,0.01].$ That is, given $0 = t_k < t_{f_{\scriptstyle i}} \le t_{k + 1} = 1,$ then $0 < t_{f_{\scriptstyle i}} \le 0.01,$ and so the conditional expectation of $t_{f_{\scriptstyle i}}$ cannot be greater than $0.01.$ But $P(\tau > t_{f_{\scriptstyle i}} - t_k \mid t_k < t_{f_{\scriptstyle i}} \le t_{k + 1}) = 1$ whenever $\tau > 0.01,$ so the value of the integral in $(1)$ is at least $\int_{0.01}^1 1\,d\tau = 0.99.$

Answer 2

The expectation can also be written as an integral. The two integrals can be transformed into each other by integration by parts.