Lower-bound for total-variation between two sub-gaussian random variables Problem setup
Let $X$ and $Y$  random variables on the real line with following properties:


*

*$X$ and $Y$ are $\sigma^2$ sub-gaussian

*$E[X] = 0$ 

*and $E[Y] = \Delta$.


Question
Whats the lower bound on total variation distance between $X$ and $Y$?
Let:
$$
\big|\big|X-Y\big|\big|_{TV} \geq f(\sigma,\Delta)
$$
What is $f(\sigma,\Delta)$? Are there any known (or trivial) lower bounds in the literature?
Few obvious properties $f(\sigma,\Delta)$ satisfy :


*

*$f$ decreases as $\sigma$ increases (fixed $\Delta$).

*$f$ increases as $\Delta$ increases (fixed $\sigma$).

 A: Sub-Gaussianity is an asymptotic property whereas expectations and the total variation are global properties.  Provided the tails of the distribution are not too heavy (and any sub-Gaussian distribution has very light tails indeed), they will have negligible effect on those global properties.  Thus we may modify any sub-Gaussian variable $X$ just a tiny bit (in probability) by mixing in a point mass far out from its mean without harming its asymptotic tail behavior.  Although this is a short total variation distance, it's an arbitrarily large difference in means.  Thus the only possible lower bound is zero.
The following is a rigorous version of this argument.  It improves on an earlier answer in that it applies more widely, requires less detailed analysis, and perhaps offers more insight into the interactions between fundamental operations (truncation and mixing) and the total variation distance.

Preliminaries
The key ideas are embodied in the properties of two operations on random variables.  To describe them, let's begin with some terminology and definitions.

*

*Let $X$ be a random variable with distribution function $F$ and let $N$ be any number.  The "tail probability" (of $X$ beyond $N$) is the chance that $X$ exceeds $N$,
$$\Pr(|X| \gt N) = \Pr(X \notin [-N,N]) = F(N) - \lim_{\varepsilon \to 0^{+}} F(-N-\varepsilon).$$
Let's call this quantity $e_F(N)$.  It equals $1$ for negative values of $N$ and eventually decreases asymptotically toward $0$ as $N$ increases.  (It is the survival function of $|X|$.)


*The total variation distance between two random variables $X$ and $Y$ is the total variation distance between their probability distributions,
$$||X-Y||_{TV} = \sup_{A\subset \mathcal{F}}\left| \Pr(X\in A) - \Pr(Y \in A) \right|.$$
$\mathcal{F}$ is the sigma-algebra of Borel sets.


*Although specific definitions appear to vary slightly, to say that $X$ is "sub-Gaussian" means its tails are no heavier than some Gaussian function $t\to C \exp(-\sigma^2 (t-\mu)^2)$.  That immediately implies sub-Gaussian variables have finite expectations, because the expectation of $|X|$ is bounded above by the integral of $C \exp(-\sigma^2 (t-\mu)^2)$, which is finite.
Truncation
Suppose $N$ is large enough to make $\Pr(X \in [-N, N])$ nonzero.  We may truncate $X$ at $N$ by removing all chance that it exceeds $N$.  The value of the new distribution function at $x$ is $0$ for $x \lt -N$, $1$ for $x \ge N$, and otherwise equals
$$F_{[N]}(x) = \frac{F(x) - \lim_{\varepsilon \to 0^{+}} F(-N-\varepsilon)}{1-e_F(N)}.$$
When we truncate $X$ at $N$, we remove a total probability of $e_F(N)$ from the tails of $X$ and, to compensate, uniformly divide all probabilities in the interval $[-N,N]$ by $1-e_F(N)$.  Therefore the most that the probability can change is $e_F(N)$ (and in fact that much change occurs for $A=[-N,N]$).  Consequently
$$||X - X_{[N]}||_{TV} = e_F(N).$$
Changing the mean
Let $0 \lt p \lt 1$ and $\Delta$ be any number, the amount by which we wish to shift the mean of $X$.  One way to accomplish this shift is to form a mixture with an atomic variable.  Its distribution function is
$$F_{\Delta,p}(x) = \cases{(1-p) F(x), \quad\quad x\lt (\Delta-\mathbb{E}(X))/p \\ (1-p)F(x) + p  \ \text{  otherwise.}}.$$
This moves a total probability of $p$, spread throughout the range of $X$, onto a single value.  When that value is $\Delta/p$ away from the mean of $X$, the mean is shifted by $\Delta$.  Let $Y$ be a random variable with this mixture distribution.  Because the probability of no event changes by any more than $p$,
$$||X - Y||_{TV} \le p.$$
Let's say that $Y$ has been attained by means of a $p$-mixture with $X$.

Solution
Let $X$ be any random variable with finite expectation.
Let $\varepsilon \gt 0$.  Pick an $N$ for which $e_F(N) \le \varepsilon/2$ and truncate $X$ at $N$.  Now shift the mean of $X_{[N]}$ to $\Delta$ using an $\varepsilon/2$-mixture. As we have seen, such a shift changes the total variation distance by at most $\varepsilon/2$.  Therefore, by the triangle inequality,
$$||X - Y||_{TV} \le ||X - X_{[N]}||_{TV} + ||X_{[N]} - Y||_{TV} \le \varepsilon/2 + \varepsilon/2 = \varepsilon.$$
We have shown that provided $X$ has finite expectation, there is always a way to truncate $X$ and shift its mean to $\Delta$, no matter what value $\Delta$ might have, without moving by more than $\varepsilon$ in the total variation distance.
These constructions put an upper bound on $|Y|$: it is no greater than the larger of $N$ or the absolute value of the position of the atom located at $2(\Delta - \mathbb{E}(X))/\varepsilon$.  Consequently the tails of $Y$ are zero, which makes them sub-Gaussian.
Since $\varepsilon$ can be arbitrarily small, the only possible lower bound on the distance is zero.

Conclusion
We have shown how to modify any finite-expectation random variable $X$ into a bounded variable $Y$ with mean $\mathbb{E}(X)+\Delta$ in such a way that the total variation distance between $X$ and $Y$ is arbitrarily small.  As a special case, when $X$ is a zero-mean $\sigma^2$ sub-Gaussian (whatever that might specifically mean), then $Y$ is $\sigma^2$ sub-Gaussian with mean $\Delta$.  This demonstrates the lower bound on $||X - Y||$ is zero, QED.
A: Here is a possible approach to a modified version of the above problem. The modification preserves the motivation of the original problem.
Modification - $X$ and $Y$ are constrained to have $\sigma^2$ variance instead of being $\sigma^2$ sub-gaussian.
Let $E[X] = -\Delta$ and $E[Y] = \Delta $ 
\begin{eqnarray}
\Big(\int_z z\big(p_X(z)- p_Y(z)\big)dz \Big)^2 &\leq& \Big(\int_z \big|z\big|\big|\;p_X(z)- p_Y(z)\;\big|dz \Big)^2\\
 &=&  \Big(\int_z \big|z\big|\sqrt{\big|\;p_X(z)- p_Y(z)\;\big|}\sqrt{\big|\;p_X(z)- p_Y(z)\;\big|}dz \Big)^2 \\
&&\leq \Big(\int_z z^2\big|\;p_X(z)- p_Y(z)\;\big|dz \int_z \big|\;p_X(z)- p_Y(z)\;\big|dz \Big)\\
&&\overset{(a)}{\leq}\Big(\int_z z^2\big(p_X(z)+p_Y(z)\big)dz \int_z \big|\;p_X(z)- p_Y(z)\;\big|dz \Big)\\
&&\overset{(b)}{\leq} \Big(2\big(\sigma^2+\Delta^2\big) \int_z \big|\;p_X(z)- p_Y(z)\;\big|dz \Big)\\
&&=\Big(2\big(\sigma^2+\Delta^2\big) 2\big|\big|\;p_X(z)- p_Y(z)\;\big|\big|_{TV}\Big)\\
\end{eqnarray}
Re-writting the above
$\big|\big|\;p_X(z)- p_Y(z)\;\big|\big|_{TV} \geq \frac{\Big(\int_z z\big(p_X(z)- p_Y(z)\big)dz \Big)^2}{4(\sigma^2+\Delta^2)} \overset{(c)}{\geq} \frac{4\Delta^2}{4(\sigma^2+\Delta^2)} {\geq} \frac{\Delta^2}{\sigma^2+\Delta^2} $
(a): $|a-b| \leq a+b$, when both $a$ and $b$ are positive.
(b): Because $Var(X), Var(Y) \leq \sigma^2 $ and  $E[X]^2 = E[Y]^2 = \Delta^2$
(c): Because $E[X] = -\Delta, E[Y] = \Delta$
Note: Solving the above problem as a convex optimization problem might lead to better bounds.
Any ideas are appreciated and thanks in advance!
