Finding $E(X\cdot\mathbb{1}_{X\ge t}\mid Y_{t})$ where $X\sim\text{Exp}(\lambda)$ and $Y_{t}=\min(t,X)$

Question

Suppose $X\sim \text{Exp}(\lambda)$ and $Y_{t}=\min(t,X)$. Show that $$E(X\cdot\mathbb{1}_{X\ge t}\mid Y_{t})=\left(t+\frac{1}{\lambda}\right)\cdot\mathbb{1}_{X\ge t}$$

I know that since $X\cdot\mathbb{1}_{X\ge t}$ and $Y_{t}$ are independent and thus $$E(X\cdot\mathbb{1}_{X\ge t}\mid Y_{t})=E(X\cdot\mathbb{1}_{X\ge t})$$ However evaluating the integral gives $$\int_t^\infty x\lambda e^{-\lambda x}\,dx=\left(t+\frac{1}{\lambda}\right)\cdot e^{-\lambda t}$$ I must be mistaken somewhere along the line. If you could give a hint or correct my mistake I would be thankful.

Answer 1

One issue: $X\cdot\mathbb 1_{X\ge t}$ and $Y_t$ are not independent: Both events $\{X\cdot\mathbb 1_{X\ge t}>t\}$ and $\{Y_t<t/2\}$ have positive probability but their intersection has probability $0$, being empty.

On the event $\{Y<t\}$ you have $X\cdot\mathbb 1_{X\ge t}=0$. The event $\{Y= t\}$ coincides with the event $=\{X\ge t\}$ on which you have $X\cdot\mathbb 1_{X\ge t}=X$. It follows that $$ E(X\cdot\mathbb 1_{X\ge t}| Y_t)= \mathbb 1_{X\ge t} E(X | X\ge t). $$