What is the point of measuring statistical distance? On pg. 378 of "Cryptography with Tamperable and Leaky
Memory", Kalai et al. claim two probability distributions are $e(k)$ close if the distance between them is at most $e(k)$.  
What is significance of two distributions X and Y being "close to" or "far from" each other? Why would anybody care, especially in cryptography?
 A: The standard game that's played in cryptographic proofs is to say:


*

*Pretend you can break (within reasonable time bounds) a cryptosystem using some procedure. Breakage here just means subverting some desirable property of the cryptosystem that you're trying to demonstrate holds. 

*Then show that using this procedure you can solve some really hard math problem that's assumed to be unsolvable by creating an instance of the cryptographic protocol that encodes the really hard math problem and whose breakage will give you the answer to the really hard math problem.

*Show that the procedure won't be able to tell the difference between a real version of the cryptographic protocol and the one you've created to get it to solve the really hard math problem.

*Conclude that since nobody knows how to solve the math problem it's unlikely that someone can break the cryptosystem.


I imagine that the distance between distributions relates to point 3. If the procedure can distinguish between the simulated problem and a real problem then your encoding of the math problem hasn't really accomplished much, since the procedure may only work on real instances.
A: Remember that continuous probability distributions can be represented analytically, as a curve in the plane. 
Suppose we have two curves represented by f(x) and g(x). One way to define the distance between them is the greatest value of the absolute value of f(x)-g(x), the distance between the two ordinates at the abscissa x.
If this value is small, then the functions are close. Otherwise, nothing is guaranteed. This forms the "statistical distance."
Often, we want to see if our guess of the true probability distribution is correct. One way of doing this is to estimate, using computational methods, the estimated density of the sample. Then, using the test above, we can see if our guess is correct.
This is just reason I can think of.
A: If you look at their definition given in [7], it says,
$\mathrm{dist}(X,Y) = \max_{A \subseteq S} \left| \mathrm{Pr}(X \in A) - \mathrm{Pr}(Y \in A)\right|$
where $X$ and $Y$ are random variables over $S$.
This is a discrete case of the total variation distance.
