# Estimating number of failures

I have a set ${X}$ of $N$ Weibull-distributed variables that are differently parameterized. Let's denote as $S_i(t)$ the probability of survival for variable $i$ at time $t$. $F_i(t) = 1 - S_i(t)$ is the failure probability for variable $i$. I'd like to estimate the expected # of total failures at time $T$. Intuitively, since the ${X}$ are independent, the probability of all ${X}$ surviving until $T$ is $\Pi_i S_i(T)$, and the probability of 1 failure in the set is then $F_{tot} = 1 - \Pi_i S_i(T)$, and the expected # of failures is then $F_{tot} * N$. But I'm not sure if my logic is quite correct. If not, what's the right approach to calculating total # of failures at a given time?

Let us look at $x_1$, without loss of generality. The probability that it fails before time $T$ is $F_1(T)$. This is the expected value of a variable that takes on the value $1$ if $x_1 < T$ and $0$ otherwise:

$$F_1(T) = \int_0^{\infty}1(x < T)f(x)dx$$

Similarly, all the other $F_i(T)$ are expected values of the appropriate indicator variable. This motivates a view of the $F_i(T)$ as expected values - the $F_i(T)$ can be interpreted as the expected number of failures of a single observation of variable $i$ before time $T$, which is just the probability of failure before time $T$.

Now, the total number of failures $n(T)$ before time $T$ can be written:

$$n(T) = \sum_{i=1}^NX_i(T)$$

where, for each fixed $T$, $X_i(T)$ is either 1 or 0 depending upon whether a failure occurred or not respectively. You can see how the values 1 or 0 match up with the indicator function values in the first equation. This is just the sum of $N$ Bernoulli variates with (possibly) different probabilities, which probabilities equal the $F_i(T)$.

The total expected number of failures before time $T$, $\mathbb{E}X(T)$, is therefore just the sum of the expected number of failures for each of the different items:

$$\mathbb{E}X(T) = \sum_i F_i(T)$$

• I don't understand your first equation: what's $f(x)$? If it's the Weibull pdf, then the equation is wrong: $F_1(T)=P(t_1\leq T)=\int_{-\infty}^Tf(x)dx$ Commented Dec 26, 2017 at 22:36
• @DeltaIV - Quite correct! I wrote everything out in terms of $S_1(t)$ and messed up converting to the failures. Thanks! Commented Dec 26, 2017 at 22:43
• ahhhh - ok, now it's clear! I'll delete my answer, yours is better. Had I simplified my expression $\mathbb{E}[f(t)]=2F(t)(1-F(t))+2F(t)^2$, I may have noticed that it corresponded to just $2F(t)-2F(t)^2+2F(t)^2=2F(t)$ , i.e., your expression, but I didn't have time...anyway, can I suggest you to add that $n(t)=\sum_{i=1}^N X_i(t)$ where for each fixed $t$ $X_1(t),\dots,X_N(t)$ are Bernoulli RVs, not identically distributed? I think this clarifies your derivation, and resembles the classical presentation of the Binomial distribution (where however the $X_i$ are iid). Commented Dec 26, 2017 at 22:56
• @DeltaIV - Good idea, I'll do that. Commented Dec 26, 2017 at 23:16
• Thanks, @jbowman, this was my first take on it, but I was somewhat put off by the fact that the cumulative hazard for weibull allowed for "repairs" and could reflect multiple failures, whereas in clinical trials an event can only occur only once. Commented Dec 27, 2017 at 16:59