Considering a system containing nine components, where the lifetimes X (in years) of the components are independent random variables and each has the probability density function:
$$f_X(x)=\begin{cases}\frac{2x}{15}& for \ \ 1 < x < 4\ \\ 0 & otherwise \end{cases}$$
Find the probability that the eighth component failure will occur within 3.5 years.
Using Order Statistics Theorem, I worked out that the probability density function of time elapsing immediately before the 8th component failure occurs to be
$$f_{Y8}(y_8)=\begin{cases}9.6y_8(\frac{y_8^2}{15})^7(1-\frac{y_8^2}{15}) & for \ \ 1 < x < 4\ \\ 0 & otherwise \end{cases}$$
However, the probability density function does not work out to be 1 and I am confused as to why.
I would be very grateful to receive any constructive feedback regarding my question. Thank you!
A route that avoids order statistics.
For $t>0$ let $N\left(t\right)$ denote the number of failures in interval $\left[0,t\right]$.
Then $N\left(t\right)$ has binomial distribution with parameters $n=9$ and $p=F\left(t\right)=\int_{-\infty}^{t}f_{X}\left(x\right)dx$.
To be found is $P\left(N\left(3.5\right)\geq8\right)$.
This agrees with $F_{X_{\left(8\right)}}\left(3.5\right)$ where $F_{X_{\left(8\right)}}$ is the CDF of order statistic $X_{\left(8\right)}$ which is:
$$F_{X_{\left(8\right)}}\left(t\right)=\sum_{j=8}^{9}\binom{9}{j}F\left(t\right)^{j}\left(1-F\left(t\right)\right)^{9-8}=9F\left(t\right)^{8}\left(1-F\left(t\right)\right)+F\left(t\right)^{9}$$(see here)
You can find the PDF of $X_{\left(8\right)}$ by differentiating the CDF (and make use of the fact that $F(t)$ differentiates to $f_X(t)$), but there is really no need for that.