Lecture 13: The Ahlswede-Winter Inequality

1. Useful versions of the Ahlswede-Winter Inequality

Theorem 1 Let {Y} be a random, symmetric, positive semi-definite {d \times d} matrix such that {{\mathrm E}[ Y ] = I}. Suppose {\lVert Y \rVert \leq R} for some fixed scalar {R \geq 1}. Let {Y_1, \ldots, Y_k} be independent copies of {Y} (i.e., independently sampled matrices with the same distribution as {Y}). For any {\epsilon \in (0,1)}, we have

\displaystyle {\mathrm{Pr}}\Bigg[ (1-\epsilon) I \:\preceq\: \frac{1}{k} \sum_{i=1}^k Y_i \:\preceq\: (1+\epsilon) I \Bigg] ~\geq~ 1 - 2d \cdot \exp( - \epsilon^2 k / 4 R ).

This event is equivalent to the sample average {\frac{1}{k} \sum_{i=1}^k Y_i} having minimum eigenvalue at least {1-\epsilon} and maximum eigenvalue at most {1+\epsilon}.

Proof: We apply the Ahlswede-Winter inequality with { X_i = \big(Y_i - {\mathrm E}[ Y_i ] \big) / R}. Note that {{\mathrm E}[ X_i ] = 0}, {\lVert X_i \rVert \leq 1}, and

\displaystyle \begin{array}{rcl} {\mathrm E}[ X_i^2 ] &=& \frac{1}{R^2} {\mathrm E}\big[ \big(Y_i - {\mathrm E}[Y_i] )^2 \big] \\ &=& \frac{1}{R^2} \Big( {\mathrm E}[ Y_i^2 ] - 2 {\mathrm E}[Y_i]^2 + {\mathrm E}[Y_i]^2 \Big) \\ &\preceq& \frac{1}{R^2} {\mathrm E}[ Y_i^2 ] \qquad(\mathrm{since}~ {\mathrm E}[ Y_i ]^2 \succeq 0)\\ &\preceq& \frac{1}{R^2} {\mathrm E}[ \lVert Y_i \rVert \cdot Y_i ] \\ &\preceq& \frac{R}{R^2} {\mathrm E}[ Y_i ] \end{array}

Finally, since {0 \preceq {\mathrm E}[Y_i] \preceq I}, we get

\displaystyle \lambda_{\mathrm{max}} \big( {\mathrm E}[X_i^2] \big) ~\leq~ 1/R. \ \ \ \ \ (1)

 

Now we use Claim 15 from the Notes on Symmetric Matrices, together with the inequalities

\displaystyle \begin{array}{rcl} 1 + x &\leq& e^x \quad\forall x \in {\mathbb R} \\ e^x &\leq& 1 + x + x^2 \quad\forall x \in [-1,1]. \end{array}

Since {\lVert X_i \rVert \leq 1}, for any {\lambda \in [0,1]}, we have { e^{\lambda X_i} \preceq I + \lambda X_i + \lambda^2 X_i^2 }, and so

\displaystyle {\mathrm E}[ e^{\lambda X_i} ] ~\preceq~ {\mathrm E}[ I + \lambda X_i + \lambda^2 X_i^2 ] ~\preceq~ I + \lambda^2 {\mathrm E}[ X_i^2 ] ~\preceq~ e^{ \lambda^2 {\mathrm E}[ X_i^2 ] }.

Thus by (1) we have

\displaystyle \lVert {\mathrm E}[ e^{\lambda X_i} ] \rVert ~\leq~ \lVert e^{ \lambda^2 {\mathrm E}[ X_i^2 ] } \rVert ~\leq~ e^{ \lambda^2 / R }.

The same analysis also shows that {\lVert {\mathrm E}[ e^{-\lambda X_i} ] \rVert \leq e^{ \lambda^2 / R }}. Substituting these two bounds into the basic Ahlswede-Winter inequality from the previous lecture, we obtain

\displaystyle {\mathrm{Pr}}\Bigg[~ \Big\lVert \sum_{i=1}^k \frac{1}{R} \big(Y_i - {\mathrm E}[Y_i] \big) \Big\rVert >t ~\Bigg] ~\leq~ 2d \cdot e^{-\lambda t} \prod_{i=1}^k e^{ \lambda^2 / R} ~=~ 2d \cdot \exp( -\lambda t + k \lambda^2 / R ).

Substituting {t = k \epsilon / R} and {\lambda = \epsilon/2} we get

\displaystyle {\mathrm{Pr}}\Bigg[~ \Big\lVert \frac{1}{R} \sum_{i=1}^k Y_i - \frac{k}{R} {\mathrm E}[Y_i] \Big\rVert > \frac{k \epsilon}{R} ~\Bigg] ~\leq~ 2d \cdot \exp( - k \epsilon^2 / 4R ).

Multiplying by {R/k} and using the fact that {{\mathrm E}[Y_i]=I}, we have bounded the probability that any eigenvalue of the sample average matrix {\sum_{i=1}^k Y_i/k} is less than {1-\epsilon} or greater than {1+\epsilon}. \Box

Corollary 2 Let {Z} be a random, symmetric, positive semi-definite {d \times d} matrix. Define {U := {\mathrm E}[ Z ]} and suppose {Z \preceq R \cdot U} for some scalar {R \geq 1}. Let {Z_1, \ldots, Z_k} be independent copies of {Z}. For any {\epsilon \in (0,1)}, we have

\displaystyle {\mathrm{Pr}}\Bigg[ (1-\epsilon) U \:\preceq\: \frac{1}{k} \sum_{i=1}^k Z_i \:\preceq\: (1+\epsilon) U \Bigg] ~\geq~ 1 - 2d \cdot \exp( - \epsilon^2 k / 4 R ).

Proof: Let {U^{+/2} := (U^+)^{1/2}} denote the square root of the pseudoinverse of {U}. Let {I_{\mathrm{im}~U}} denote the orthogonal projection on the image of {U}. Define the random, positive semi-definite matrices

\displaystyle Y ~:=~ U^{+/2} \cdot Z \cdot U^{+/2} \qquad\mathrm{and}\qquad Y_i ~:=~ U^{+/2} \cdot Z_i \cdot U^{+/2}.

Because {Z_i \succeq 0} and {U = {\mathrm E}[\sum_i Z_i]}, we have {\mathrm{im}(Z_i) \subseteq \mathrm{im}(U)}. So Claim 16 in Notes on Symmetric Matrices implies

\displaystyle (1-\epsilon) U \:\preceq\: \frac{1}{k} \sum_{i=1}^k Z_i \:\preceq\: (1+\epsilon) U \qquad\Longleftrightarrow\qquad (1-\epsilon) I_{\mathrm{im}~U} \:\preceq\: \frac{1}{k} \sum_{i=1}^k Y_i \:\preceq\: (1+\epsilon) I_{\mathrm{im}~U}.

We would like to use Theorem 1 to obtain our desired bound. We just need to check that the hypotheses of the theorem are satisfied. By Fact 6 from the Notes on Symmetric Matrices, we have

\displaystyle Y ~=~ U^{+/2} \cdot Z \cdot U^{+/2} ~\preceq~ U^{+/2} \cdot (R \cdot U) \cdot U^{+/2} ~=~ R \cdot I_{\mathrm{im}~U},

showing that {\lVert Y \rVert \leq R}. Next,

\displaystyle {\mathrm E}[Y] ~=~ U^{+/2} \cdot {\mathrm E}[Z] \cdot U^{+/2} ~=~ U^{+/2} \cdot U \cdot U^{+/2} ~=~ I_{\mathrm{im}~U}.

So the hypotheses of Theorem 1 are almost satisfied, with the small issue that {{\mathrm E}[Y]} is not actually the identity, but merely the identity on the image of {U}. But, one may check that the proof of Theorem 1 still goes through as long as every eigenvalue of {{\mathrm E}[Y]} is either {0} or {1}, i.e., {{\mathrm E}[Y]} is an orthogonal projection matrix. The details are left as an exercise. \Box

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One Response to Lecture 13: The Ahlswede-Winter Inequality

  1. Pingback: Lecture 15: Low-rank approximation of matrices | UBC CPSC 536N

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