I am continuing to try and fully understand the argument why Matrix inversion is BQP-complete according to the proof given in the HHL paper here, and I have hit another snag.
In this question here, I got clarification about the operator $U$ so-defined, where eventually we get for $A = I - U^{-1/T}$.
$A^{-1} = \sum_{k \geq 0} U^k e^{-k/T}$ where $U^{3T} = I$. The author then writes
This (applying $A^{-1}$) can be interpreted as applying $U^t$ for $t$ a geometrically-distributed random variable. [...]
I'm not sure why this statement is true.
From what I understand, $U$ is an operator that operates on two registers, one of size $3T$ where $T$ is the number of gates, and the other register of size $n$ initialized to $|0 \rangle ^{\otimes n}$. $A^{-1}$ is a matrix consisting of a convergent series of powers of $U$ where the coefficients are exponentials. As $k \rightarrow \infty$, the powers of $U$ are divided by such large exponentials that they vanish. I have reasoned why $A^{-1}$ is given by this sum since by left/right multiplying it by $A$ we get a telescoping series where the only value of $k$ that doesn't vanish is $k=0$.
The point of all this is that if we apply $U^t$ for $T+1 \leq t \leq 2T$, then we will be left with a state $|t+1 \rangle U_1 \ldots U_T |0 \rangle ^{\otimes n}$, where the second register will correspond to applying the $T$ gates to our initial state. I do not understand the portion of this argument consisting of interpreting the application of $A^{-1}$ as a geometric random variable.
Moreover the author/s state that measuring $t$ in the range $T+1 \leq t \leq 2T$ occurs with probability $\frac {e^{-2}}{1+e^{-2}+e^{-4}}$, and I am not sure how they got this. Any hints appreciated in order to understand this computation.
Edit: Working on this I find that the probability of measuring $T+1 \leq t \leq 2T$ should be something like :
$(\sum_{k=T+1}^{2T}(\sum_{j=0}^{\infty}e^{-(k+3j)/T}) )/ (\sum_{n=0}^{\infty}e^{-n/T}) = (\sum_{k=T+1}^{2T}e^{-k/T}(\frac{1}{1-e^{-3/T}}))/(\frac{1}{1-e^{-1/T}}) = (\frac{1}{1-e^{-3/T}})/(\frac{1}{1-e^{-1/T}}) \sum_{j=0}^{T-1} e^{-(j+T+1)/T} = (\frac{1}{1-e^{-3/T}})/(\frac{1}{1-e^{-1/T}}) e^{-(T+1)/T}(\frac{1 - e^{-1}}{1-e^{-1/T}})$
Thanks!