Expectation and Variance of a Random Variable

Question

let the variable be

X = $\int_{0}^1 I_{(B_t>0)}dB_t$

I am required to calculate the expectation and variance. $B_{t}$ is standard brownian motion.

My Approach :

Approach 1 : $B_{t}$ is within 0 to 1 in limits, so indicator function will be 1. Thus X is basically B(1). Hence, Mean 0, Variance 1.

Approach 2 : Using Ito isometry, take variance of X is expectation of RHS squared. Square of indicator function is same as itself, $dB_{t}^2$ is $dt^2$ and E($I_{B_t>0}) = 1/2$, so variance of X is 1/2.

Why do the different approaches yield different results(variance of X)? Which one is correct, if at all?

Answer 1

Regarding Approach 1, $t \in (0,1)$ does not imply $B_t \in (0,1)$. The variance calculation in Approach 2 is correct though. For the expected value, note that $$\mathbb{E}[X] = \mathbb{E}[\mathbb{E}[X|\mathcal{F}_t]] = \mathbb{E}\left[\int_0^1 I_{B_t>0}\mathbb{E}[dB_t|\mathcal{F}_t]\right]=0.$$