Conditional expectation and LOTUS to prove an identity for $\mathbb E[B(X)\mid X>k]$.

Question

Let $X$ be a continuous and real random variable, with density $f(x)$, and let $X\in[0,w]$. Let $B(x)$ be a measurable real-valued and let $k\in[0,w]$ be a known scalar. Is it true that: \begin{equation} E[B(X)|X>k]=\frac{1}{1-F(k)}\int_{k}^{w}B(x)f(x)dx \end{equation}

Thanks.

Answer 1

The following equation is by definition $$E[B(x) | \{ X > k \}] = \int_{x = 0}^{\omega} B(x) f_{X|\{X > k \}}(x) dx$$

where $f_{X | \{X > k \}}(x)$ is the conditional distribution of $X$ conditioned on the event $\{X > k \}$ which is just equal, also by definition, to $\frac{f_X(x)}{P(X > k)} = \frac{f_X(x)}{1 - P(X < k)}$ for $k < x < \omega$.

$$f_{X | \{X > k \}}(x) = \begin{cases} \frac{f_X(x)}{1 - P(X < k)} & k < x < \omega\\ 0 & \text{otherwise} \end{cases}$$

Therefore,

$$E[B(x) | \{ X > k \}] = \int_{x = k}^{\omega} B(x) \frac{f_X(x)}{1 - P(X < k)} dx$$