This is so trivial that authors usually don't bother to give an explicit proof. But for me there is some vagueness.

We say that two ensembles $X_n$ and $Y_n$ are statistically close, if $$ \Delta(n) = 1/2 \sum_{\alpha}|\mathbb{P}[X_n = \alpha] - \mathbb{P}[Y_n = \alpha]| $$ is negligible in n. The probability is taken over the randomness of $X_n$ and $Y_n$ respectively.

We say that two ensembles are computationally indistinguishable if for every PPT-adversary D we have $$ |\mathbb{P}[D(X_n) \to 1] - \mathbb{P}[D(Y_n) \to 1]|$$ is negligible.

Why the former implies the latter?

I do understand that for every deterministic function $f$ we have $\Delta(f(X), f(Y)) \le \Delta(X, Y),$ where $\Delta(\cdot, \cdot)$ is the statistical distance.

But in the case of PPT adversaries $D$ is not deterministic, there are implicit random coins. Why can we treat PPT-algorithm $D$ as deterministic function?


2 Answers 2


A probabilistic distinguisher is still a deterministic function of its input and random coins. So a probabilistic distinguisher trying to distinguish $X$ from $Y$ is equivalent to a deterministic distinguisher trying to distinguish $(X,R)$ from $(Y,R)$ where $R$ is a uniform distribution over random coins (importantly: independent of $X$/$Y$).


\begin{align} \Delta\Bigl( (X,R), (Y,R) \Bigr) &= \frac12 \sum_{\alpha,r} \Bigl| \Pr[(X,R)=(\alpha,r)] - \Pr[(Y,R)=(\alpha,r)]\Bigr| \\ &= \frac12 \sum_{\alpha,r} \Bigl| \Pr[X=\alpha]\Pr[R=r] - \Pr[Y=\alpha]\Pr[R=r]\Bigr| \\ &= \frac12 \sum_{\alpha,r} \Bigl| \Pr[X=\alpha] - \Pr[Y=\alpha]\Bigr| \Pr[R=r] \\ &= \frac12 \sum_{\alpha} \Bigl| \Pr[X=\alpha] - \Pr[Y=\alpha]\Bigr| \;\underbrace{\sum_r \Pr[R=r]}_{=1} \\ &= \frac12 \sum_{\alpha} \Bigl| \Pr[X=\alpha] - \Pr[Y=\alpha]\Bigr| \\ &= \Delta(X,Y) \end{align}

In short, having access to some distribution that is independent of $X$/$Y$ doesn't help (or hurt) to distinguish $X$ from $Y$.


Another way to see this would be to try and upper bound the distinguishing advantage for any distinguisher and relate that to the statistical distance.


Since the following answer is really good, I will just give ideas without proofs.

Was supposed to be :

Since @Mikero's answer is really good...

What happens when you answer late and do not proof-read: self-Facepalm and hides in shame for bragging about my answer

Let $(X, Y)$ be two random variables on the set $\mathcal{X}$. We denote by $\Delta^D(X;Y)$ the distinguishing advantage of a distinguisher $D$ with binary output and by $\delta(X,Y)$ by the maximum distinguishing advantage for $(X,Y)$.(i.e the advantage of one optimal distinguisher).

We need to do two things:

  • Give an "explicit description" of a deterministic distinguisher $\mathcal{D}$ that has advantage $\delta(X;Y)$
  • show that $\delta(X;Y) = \Delta(X;Y)$
  • The conclusion will be the implication in the question

First we show an explicit optimal deterministic distinguisher

For $X$ with distribution $Pr_X[x], x \in \mathcal{X}$ and $Y$ with distribution $Pr_Y[x]$, intuitively an optimal deterministic distinguisher $\mathcal{D}(\cdot)$ would do the following:

  • $\mathcal{D}(x) = 0$ if $Pr_X[x] \geq Pr_Y[x]$
  • $\mathcal{D}(x) = 1$, otherwise

Let $\mathcal{X}^* = \{x: Pr_X[x] \geq Pr_Y[x]\}$, we can show that $\Delta^{\mathcal{D}}(X,Y) = Pr[Y \in \mathcal{X}^*] - Pr[Y \in \mathcal{X}^*]$.

One can show that $\Delta^{\mathcal{D}}(X;y) = Pr[Y \in \mathcal{X}^*] - Pr[Y \in \mathcal{X}^*] = \delta(X;Y)$

Second, we relate the distinguishing advantage to the statistical distance

We have the following $\forall D, \Delta^D(X;Y) \leq \delta(X;Y)$ by defition, and on the other hand $\delta(X;Y) = \Delta(X;Y)$ therefore we have the following $$\forall D, \Delta^D(X;Y) \leq \Delta(X,Y)$$.

In conclusion the statistical distance gives an upper bound on the performance of any distinguisher, probabilistic included.

  • $\begingroup$ So, does it not matter if the distinguisher runs in polynomial or exponential time? $\endgroup$ Sep 20, 2019 at 8:17
  • $\begingroup$ @Hilder indeed in this case, I did not restrict the distinguisher to a particular class(PPT) of di to make the argument. An naturally more interesting case would to try an workout the advantage for a particular class, which in most case is generally not easy. $\endgroup$ Sep 25, 2019 at 7:21
  • $\begingroup$ With my way of proof I also get the answer that it does not matter how strong the distinguisher is, as long as the ensembles are statistically close then the advantage stays negligible. $\endgroup$
    – killertoge
    Nov 14, 2022 at 15:29

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