What are the decay products of HTO?
I don’t consider your working to be arbitrary at all. Let’s first consider the &#x3b2; decay including the associated $\ce{OH}$ part of $\ce{HTO}$:

$$\ce{_1^3THO -&gt; _2^3He^2+ + OH- + e-^ + \bar{\nu_e}}\tag{1}$$

As you suggested, the electron created in equation (1) won’t stick around, it is emitted into the environment, leaving us with a very reactive $\ce{He^2+}$ ion. Obviously, that is a strong oxidant as it wants to get reduced itself. The closest electron sources are $\ce{OH-}$ and $\ce{HTO}$, so we can consider a number of other reactions. However, it makes sense to organise it in a simple redox reaction. The reduced component is obviously going to be $\ce{He^2+}$ which only leaves $\ce{HTO}$ being the oxidised component — and since both hydrogen and tritium are in the $\mathrm{+I}$ oxidation state in $\ce{HTO}$, it must be oxygen which is oxidised to either peroxide or elemental oxygen. Indeed, peroxides are not stable and typically disproportionate into oxygen and water anyway, so we might as well just consider the formation of oxygen. Here goes:

$$\textbf{Reduction:}\\
\begin{gather}\ce{He^2+ + 2 e- -&gt; He^{\pm 0}}\tag{2}\end{gather}$$
$$\textbf{Oxidation:}\\
\begin{align}\ce{2HTO^{-II} &amp;-&gt; O2^{\pm0} + 4 e-}\tag{3.1}\\
\ce{2 HTO &amp; -&gt; O2 + 4 e- + 2 H+ + 2 T+}\tag{3.2}\end{align}$$
$$\textbf{Redox:}\\
\begin{gather}\ce{2 He^2+ + 2 HTO -&gt; 2 He + O2 + 2 H+ + 2 T+}\tag{4}\end{gather}$$

Equation (3) and (4) would be similar if I had chosen to oxidise only to $\ce{H2O2}$ and then would have been followed by an equation (5) to show the disproportionation of hydrogen peroxide:

$$\ce{2 H2O2 -&gt; O2 + 2 H2O}\tag{5}$$

In the end, you will arrive more or less at the same substances.

Note that to end up with hydrogen gas, you would have needed a reductive environment capable of reducing $\ce{H^{+I}}$ to $\ce{H2^{\pm0}}$ — not the case if you have a severly electron-desiring ion like $\ce{He^2+}$ around.