That's an unnecessarily tedious way to do it if you just want the variance of Y, but yeah, to find the expectation of that quadratic quantity in X, if we're given the PDF or PMF, we can easily find it using the Law of the Unconscious Statistician (LOTUS).
Note that to get E[T^2], we're integrating x^2 times the PDF, not x times the PDF.
Yes I know that but after simplification you end up with the integral i said
Thats not the final integral$, the PDF is $f(t) = \lambda e^{-\lambda t}, t \geq 0$. So $\mathbb{E}\left[T^2\right] = \int _0 ^\infty t^2 \cdot \lambda e^{-\lambda t}\,\mathrm{d}t$.$)
Im talking about when you apply IBP twicehahahahah nevermind... I hope the maths department can forgive me for accusing them of something they didn't do... thanks = \lambda \vec{v}$, where $\lambda$ is the eigenvalue for eigenvector $\vec{v}$ (this is precisely what it means for a non-zero vector $\vec{v}$ to be an eigenvector corresponding to eigenvalue $\lambda$). If you meant something else, please clarify.$)
Peter Brown will hear about this shortly. Prepare for your expulsion.
pls... Peter Brown loves me
There's nothing really universal, these are mainly rules of thumb. If in doubt, I guess just use the continuity correction (with "1/2" as the epsilon). The main thing is to make sure you do the correction the right way. The thing you originally wrote (P(X < 12) being approximated with the continuous X's P(X < 11.5)) isn't the right way, it'd have to be P(X < 12.5) to do the continuity correction. Just visualise the graphs as I explained.
Reason is due to which part of the graph you miss out on by using the continuous approximation. If X* is the continuous approximtion r.v. to X, then clearly, to approximate P(X > a), we need to use the continuity correction P(X* > a – 0.5), whereas for P(X < a), it'd be P(X* < a + 0.5). Don't need to guess.