I have the following question in Given a random sequence $X_t, t\in[1,T]$, with $X_t = \varepsilon_t$ for $t=1,\ldots,d$ and $X_t = \mu + \varepsilon_t$ for $t=d+1,\ldots,T$ and $\mu>0,\varepsilon_t\sim\mathcal{N}(0,1)$.
I now define the function $$f(b,X)=\frac{\frac{\sum_{t=1}^b(X_t)}{b}-\frac{\sum_{t=b+1}^T(X_t)}{T-b}}{\sqrt{\frac{T}{b(T-b)}}}, b=1,\ldots,T-1.$$ The expression under the fraction is the root of the variance of the expression over the fraction. Thus for any $b$ $$f(b,X)\sim\mathcal{N}(0,1).$$ I also know the maximum of $\mathbb{E}|f(b,X)|$ is assumed at $d$ with $$ \mathbb{E}\left[|f(b,X)|b=d|\right]=\mu\sqrt{\frac{d(T-d)}{T}} $$ and $\mathbb{E}|f(b,X)|$ is strictly monotonically increasing up to $d$ with $$ \mathbb{E}\left[|f(b,X)|b<d|\right]=\mu\sqrt{\frac{b}{T(T-b)}}(T-d) $$ and strictly monotonically decreasing from $d+1$ on with $$ \mathbb{E}\left[|f(b,X)|b>d|\right]=\mu\sqrt{\frac{(T-b)}{Tb}}(d). $$
How could I know establish, that $$ \mathbb{E}\left[argmax_{1\leq b < T}|f(b,X)|\right] = d =argmax_{1\leq b < T}\mathbb{E}|f(b,X)| $$