Am I computing this conditional expectation correctly?

Question

Let $f_X(t) = (5/6 - t^2)\mathsf 1_{(-1,1)}(t)$ be the density of $X$. Then according to this post, we may compute $$ \mathbb E[X\mid X>k] = \frac1{1-F(k)}\int_k^1 tf_X(t)\ \mathsf dt $$ We have $1-F(k) = \int_k^1 (5/6 - t^2)\ \mathsf dt = \frac{1}{6} \left(2 k^3-5 k+3\right)$ and $$ \int_k^1 t(5/6-t^2)\ \mathsf dt = \frac{1}{12} \left(3 k^4-5 k^2+2\right), $$ hence $$ \mathbb E[X\mid X>k] = \frac{\frac{1}{12} \left(3 k^4-5 k^2+2\right)}{\frac{1}{6} \left(2 k^3-5 k+3\right)}, $$ so that $$ \mathbb E[X\mid X>k] = \frac{1}{2} \left(2 k^3-5 k+3\right) \left(3 k^4-5 k^2+2\right). $$

Answer 1

There is no $\omega$ in your question.. but there is one in the other post, you formula should be :

$$E[X\mid X>k] = \frac1{1-F(k)}\int_k^\infty tf_X(t)\ \mathsf dt$$

Let $-1 \leq k \leq 1$ (the other cases are trivial)

$$F(k)=P(X<k)=\int_{-\infty}^k f_X(t)\ \mathsf dt=\frac{5}{6}(k+1)-\left[\frac{k^3}{3}+\frac{1}{3}\right]$$

$$E\left[X 1_{X>k}\right]=\int_{-\infty}^\infty tf_X(t)1_{t>k}\ \mathsf dt=\int_{k}^1 {t\left(\frac{5}{6}-t^2\right)}\ \mathsf dt=\frac{5}{6}\left(\frac{1}{2}-\frac{k^2}{2}\right)-\left[\frac{1}{4}-\frac{k^4}{4}\right]$$

You can finish by using the following formula $$E\left[X |X>k\right]=\frac{E\left[X 1_{X>k}\right]}{P(X>k)}$$