Changing expectations into probability. 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.
 A: 
$\tau$ 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.

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 
$\int_{0.01}^1 1\,d\tau = 0.99.$
A: The expectation can also be written as an integral. The two integrals can be transformed into each other by integration by parts.
