# Count Sketch probability bound

I have been reading up on the Count Sketch algorithm, and I stumpled upon the Count Sketh algorithm explained in section 5 of https://www.cs.dartmouth.edu/~ac/Teach/data-streams-lecnotes.pdf. Then, I tried to solve some of the exercises. However, exercise 5-3 is causing me some problems.

If we let $$\hat{f}_a$$ denote the estimated frequency associated with a key $$a$$ and $$f_a$$ be actual frequency associated with key $$a$$, then it can be show that $$E[\hat{f}_a]=f_a$$. Furthermore, for $$j\in[n]$$, we let $$\mathbf{f}_{-j}$$ be the $$(n-1)$$ dimensional vector obtained by dropping the $$i$$th entry of frequency vector $$\mathbf{f}.$$ It can then be shown that $$\text{Var}[\hat{f}_a]=\frac{||\mathbf{f}_{-a}||_2^2}{k}$$ when we use the hash function $$h : [n] \rightarrow [k]$$. Thus, using Chebyshev's inequality, it can be shown that $$\text{Pr}[|\hat{f}_a - f_a| \geq \epsilon ||\mathbf{f}_{-a}||_2] \leq \frac{1}{3}$$ for $$k=\frac{3}{\epsilon^2}$$.

In exercise 5-3, $$\textbf{f}_{-a}^{\text{res}(\ell)}$$ denotes the $$(n-1)$$-dimensional vector obtained by setting the $$\ell$$ largest (by absolute value) entries to zero in $$\mathbf{f}_{-a}$$. The exercise then essentially asks the reader to show that $$\text{Pr}[|\hat{f}_a - f_a| \geq \epsilon ||\mathbf{f}_{-a}^{\text{res}(\ell)}||_2] \leq \frac{1}{3}$$ where $$k=\frac{6}{\epsilon^2}$$ and $$\ell=1/\epsilon^2$$. However, I don't know how to show that the inequality above holds. I have tried using Chebyshev's inequality, but I still can't seem to make it work. Any help would be appreciated.

$$\newcommand{\norm}{\left\lVert#1\right\rVert}$$
Denote the support of $$f$$ by $$f_1\le f_2\le...\le f_s$$. Given a frequency sequence $$f$$, denote by $$E_{f,a}$$ the random variable $$\hat{f}_a$$ where the stream frequencies are given by $$f$$ (the evaluation depends on the inner randomness of the sketch and the input stream).
The probability that the hash of one of the top $$\ell=1/\epsilon^2$$ elements collides with the hash of some given element $$a$$ is bounded by $$\frac{\ell}{k}$$ (union bound). Conditioning on the event that such a collision did not occur, the evaluation $$\hat{f}_a$$ does not depend on the top $$\ell$$ entries, as they do not affect the hash entry of $$a$$. More formally, conditioned on the event $$C=\bigcap\limits_{i=0}^{\ell-1}\mathbb{1}_{h\left(f_{s-i}\right)\neq h(a)}$$, $$E_{f,a}$$ and $$E_{f^{res(\ell)},a}$$ are equally distributed. This yields:
$$\Pr\left[\left|\hat{f}_a-f_a\right|\ge \epsilon\norm{f_{-a}^{res(\ell)}}\right]= \Pr[C]\cdot \Pr\left[\left|\hat{f}_a-f_a\right|\ge \epsilon\norm{f_{-a}^{res(\ell)}}\big| C\right]+\\ \left(1-\Pr[C]\right)\Pr\left[\left|\hat{f}_a-f_a\right|\ge \epsilon\norm{f_{-a}^{res(\ell)}} \big|\overline{C}\right]\le\\ \Pr\left[\left|\hat{f}_a-f_a\right|\ge \epsilon\norm{f_{-a}^{res(\ell)}}\big| C\right]+\frac{1}{6}$$
Since $$E_{f,a}, E_{f^{res(\ell)},a}$$ are equally distributed (conditioned on $$C$$), the left hand term equals $$\Pr\left[\left|\hat{f}_a-f_a\right|\ge \epsilon\norm{f_{-a}}\right]$$, which you can bound by $$1/6$$ with your original result.