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I think I need some help with this problem.

According to p.544 of Feller's An Introduction To Probability Theory. Vol II if $|a_k|\leq c_k$, $|b_k|\leq c_k$ then we have $|\prod_{i=1}^na_i -\prod_{i=1}^n b_i|\leq\sum_{k=1}^n (a_k\prod_{i=1}^{k-1}c_i-b_k\prod_{i=k+1}^n c_i)$

I could only prove $|a^n-b^n|=\bigg|\sum_{i=0}^{n-1}(a^{i+1}b^{n-(i+1)}-a^ib^{n-i})\bigg|=\bigg|\sum_{i=0}^{n-1}a^{i+1}b^{n-1-i}-\sum_{i=0}^{n-1}a^{i}b^{n-i}\bigg|\\=\bigg|(a-b)\sum_{i=0}^{n-1}a^ib^{n-1-i}\bigg|\leq n|a-b|c^{n-1}$

but I don't know how I can do the more general case.

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  • $\begingroup$ What does $\sum_{k=1}^n c_1\cdot ... \cdot c_{k-1}a_k-b_kc_{k+1}\cdot ...\cdot c_n$ mean? Do you mean $\prod_{k=1}^n c_ka_k-\prod_{k=1}^nc_kb_k$? $\endgroup$ – R. Burton May 22 at 19:10
  • $\begingroup$ @R.Burton I changed it, hope its more clear now $\endgroup$ – Testname420 May 22 at 19:23
  • $\begingroup$ You may want to double check your formulas. $\endgroup$ – R. Burton May 22 at 19:39
  • $\begingroup$ The inequality is used on p. 544 in An Introduction to Probability Theory and Its Applications Vol2 from Feller $\endgroup$ – Testname420 May 22 at 19:41
  • $\begingroup$ Aah, that helps a lot. $\endgroup$ – R. Burton May 22 at 19:47
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Assuming that the line "$|\alpha_1\cdots\alpha_n-\beta_1\cdots\beta_n|\leq\sum_{k=1}^n\gamma_1\cdots\gamma_{k-1}\alpha_k-\beta_k\gamma_{k+1}\cdots\gamma_n$" is indeed a mistake, and that it is meant to read $|\alpha_1\cdots\alpha_n-\beta_1\cdots\beta_n|\leq\sum_{k=1}^n|\gamma_1\cdots\gamma_{k-1}\alpha_k-\beta_k\gamma_{k+1}\cdots\gamma_n|$, we have...

$$\forall k,n\in\Bbb{N}.|a_k|\leq c_k\land |b_k|\leq c_k\implies\left\vert\prod_{i=1}^na_i -\prod_{i=1}^n b_i\right\vert\leq\sum_{k=1}^n \left\vert a_k\prod_{i=1}^{k-1}c_i-b_k\prod_{i=k+1}^n c_i\right\vert$$

Now, from this post, we know that...

$$\forall\mathbf{a},\mathbf{b}\in\Bbb{C}^n:\left(\forall k\leq n\in\Bbb{N}.|a_k|\leq 1\land |b_k|\leq 1\right).\left\vert\prod_{i=1}^na_i -\prod_{i=1}^n b_i\right\vert\leq\sum_{k=1}^n \left\vert a_k-b_k\right\vert$$

That $\forall k\leq n\in\Bbb{N}.|a_k|\leq 1\land |b_k|\leq 1$ is implied by context (if not, then disregard this answer), so the question becomes "how does Feller's sum differ?"

From $|\cdot|:\Bbb{C}\to[0,\infty)$, we know that $\forall k\in\Bbb{N}.|a_k|,|b_k|\geq 0$, whence...

$$\forall k\in\Bbb{N}.0\leq|a_k|,|b_k|\land|a_k|,|b_k|\leq c_k\implies 0\leq c_k$$

Thus, $\prod_{k\in K}c_k\geq0$ for all valid indexing sets $K$. Saying anything more than this will require more information on what $\mathbf{c}$ is. Suffice to say that if...

$$\exists x\in[1,\infty).\prod_{i=1}^{k-1}c_i=\prod_{i=k+1}^n c_i=x$$

...then...

$$\sum_{k=1}^n \left\vert a_k-b_k\right\vert\leq\sum_{k=1}^n \left\vert a_k\prod_{i=1}^{k-1}c_i-b_k\prod_{i=k+1}^n c_i\right\vert=x\sum_{k=1}^n \left\vert a_k-b_k\right\vert$$

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  • $\begingroup$ Thanks you for your hard work $\endgroup$ – Testname420 May 24 at 15:57

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