Return to Answer

I would like indeed to improve the lower bound on $c$, to about $0.032$. I'll be using the same notations as in Anthony Quas's answer. As he showed using the result by George Lowther, $$(*)\qquad P(|S|\ge1)\ge\frac{2a^2}{3(1+a)^2},$$ which actually holds for all possible values of $a$ (in $[0,1]$).

The main new ingredient is the following lower bound on $ES^4$:
$$(**)\qquad ES^4=3\Big(\sum_i a_i^2\Big)^2-2\sum_i a_i^4\ge3-2a^2, $$ which suggests that $|S|^2$ must be "large enough with a large enough probability" (as compared with $E|S|^2=1$). On the other hand, $$ ES^4=ES^4\,\mathrm{I}\{|S|<1\}+ES^4\,\mathrm{I}\{|S|\ge1\}\le1 +\sqrt{ES^8\,P(|S|\ge1)}, $$ by the Cauchy--Schwarz inequality; here $\mathrm{I}\{\cdot\}$ is the indicator function. Therefore and in view of $(**)$, $$(***)\qquad P(|S|\ge1)\ge\frac{(2-2a^2)^2}{ES^8}. $$ By the Whittle inequality Whittle, $ES^8\le EZ^8=105$, where $Z$ is a standard normal random variable. So, combining $(*)$ and $(***)$, one has $$ P(|S|\ge1)\ge\min_{a\in[0,1]}\Big(\frac{2a^2}{3(1+a)^2}\bigvee\frac{(2-2a^2)^2}{105}\Big)>0.032. $$

Addendum: As shown in Rademacher-lower, the lower bound $\frac{(2-2a^2)^2}{105}$ on $P(|S|\ge1)$ can be improved to $$\frac{2(1-a^2)^3}{(6+a^2)(5+2 a^2)}.$$ Thus, the lower bound $0.032$ gets improved to $0.043$.

Addendum 2: Instead of the Paley--Zygmund inequality, used by George Lowther to show that $P(|S|>u)\ge(1-u^2)^2/3$ for $u\in(0,1)$, one can use the Cantelli inequality to get $$P(|S|>u)=1-P(1-S^2\ge1-u^2)\ge\frac{(1-u^2)^2}{(1-u^2)^2+E(1-S^2)^2} \ge\frac{(1-u^2)^2}{(1-u^2)^2+2}, $$ since $E(1-S^2)^2=ES^4-1\le2$. Now proceeding as before, one has the lower bound $$\frac{a^2}{(1+a)^2+2a^2} $$ in place of $\frac{2a^2}{3(1+a)^2}$, which results in the improvement of the lower bound on $P(|S|\ge1)$ from $0.043$ to $0.04789$.

I would like indeed to improve the lower bound on $c$, to about $0.032$. I'll be using the same notations as in Anthony Quas's answer. As he showed using the result by George Lowther, $$(*)\qquad P(|S|\ge1)\ge\frac{2a^2}{3(1+a)^2},$$ which actually holds for all possible values of $a$ (in $[0,1]$).

The main new ingredient is the following lower bound on $ES^4$:
$$(**)\qquad ES^4=3\Big(\sum_i a_i^2\Big)^2-2\sum_i a_i^4\ge3-2a^2, $$ which suggests that $|S|^2$ must be "large enough with a large enough probability" (as compared with $E|S|^2=1$). On the other hand, $$ ES^4=ES^4\,\mathrm{I}\{|S|<1\}+ES^4\,\mathrm{I}\{|S|\ge1\}\le1 +\sqrt{ES^8\,P(|S|\ge1)}, $$ by the Cauchy--Schwarz inequality; here $\mathrm{I}\{\cdot\}$ is the indicator function. Therefore and in view of $(**)$, $$(***)\qquad P(|S|\ge1)\ge\frac{(2-2a^2)^2}{ES^8}. $$ By the Whittle inequality Whittle, $ES^8\le EZ^8=105$, where $Z$ is a standard normal random variable. So, combining $(*)$ and $(***)$, one has $$ P(|S|\ge1)\ge\min_{a\in[0,1]}\Big(\frac{2a^2}{3(1+a)^2}\bigvee\frac{(2-2a^2)^2}{105}\Big)>0.032. $$

Addendum: As shown in radem-lower, the lower bound $\frac{(2-2a^2)^2}{105}$ on $P(|S|\ge1)$ can be improved to $$\frac{2(1-a^2)^3}{(6+a^2)(5+2 a^2)}.$$ Thus, the lower bound $0.032$ gets improved to $0.043$.

Addendum 2: Instead of the Paley--Zygmund inequality, used by George Lowther to show that $P(|S|>u)\ge(1-u^2)^2/3$ for $u\in(0,1)$, one can use the Cantelli inequality to get $$P(|S|>u)=1-P(1-S^2\ge1-u^2)\ge\frac{(1-u^2)^2}{(1-u^2)^2+E(1-S^2)^2} \ge\frac{(1-u^2)^2}{(1-u^2)^2+2}, $$ since $E(1-S^2)^2=ES^4-1\le2$. Now proceeding as before, one has the lower bound $$\frac{a^2}{(1+a)^2+2a^2} $$ in place of $\frac{2a^2}{3(1+a)^2}$, which results in the improvement of the lower bound on $P(|S|\ge1)$ from $0.043$ to $0.04789$.

I would like indeed to improve the lower bound on $c$, to about $0.032$. I'll be using the same notations as in Anthony Quas's answer. As he showed using the result by George Lowther, $$(*)\qquad P(|S|\ge1)\ge\frac{2a^2}{3(1+a)^2},$$ which actually holds for all possible values of $a$ (in $[0,1]$).

The main new ingredient is the following lower bound on $ES^4$:
$$(**)\qquad ES^4=3\Big(\sum_i a_i^2\Big)^2-2\sum_i a_i^4\ge3-2a^2, $$ which suggests that $|S|^2$ must be "large enough with a large enough probability" (as compared with $E|S|^2=1$). On the other hand, $$ ES^4=ES^4\,\mathrm{I}\{|S|<1\}+ES^4\,\mathrm{I}\{|S|\ge1\}\le1 +\sqrt{ES^8\,P(|S|\ge1)}, $$ by the Cauchy--Schwarz inequality; here $\mathrm{I}\{\cdot\}$ is the indicator function. Therefore and in view of $(**)$, $$(***)\qquad P(|S|\ge1)\ge\frac{(2-2a^2)^2}{ES^8}. $$ By the Whittle inequality Whittle, $ES^8\le EZ^8=105$, where $Z$ is a standard normal random variable. So, combining $(*)$ and $(***)$, one has $$ P(|S|\ge1)\ge\min_{a\in[0,1]}\Big(\frac{2a^2}{3(1+a)^2}\bigvee\frac{(2-2a^2)^2}{105}\Big)>0.032. $$

Addendum: As shown in Rademacher-lower, the lower bound $\frac{(2-2a^2)^2}{105}$ on $P(|S|\ge1)$ can be improved to $$\frac{2(1-a^2)^3}{(6+a^2)(5+2 a^2)}.$$ Thus, the lower bound $0.032$ gets improved to $0.043$.

Addendum 2: Instead of the Paley--Zygmund inequality, used by George Lowther to show that $P(|S|>u)\ge(1-u^2)^2/3$ for $u\in(0,1)$, one can use the Cantelli inequality to get $$P(|S|>u)=1-P(1-S^2\ge1-u^2)\ge\frac{(1-u^2)^2}{(1-u^2)^2+E(1-S^2)^2} \ge\frac{(1-u^2)^2}{(1-u^2)^2+2}, $$ since $E(1-S^2)^2=ES^4-1\le2$. Now proceeding as before, one has the lower bound $$\frac{a^2}{(1+a)^2+2a^2} $$ in place of $\frac{2a^2}{3(1+a)^2}$, which results in the improvement of the lower bound on $P(|S|\ge1)$ from $0.043$ to $0.04789$.

I would like indeed to improve the lower bound on $c$, to about $0.032$. I'll be using the same notations as in Anthony Quas's answer. As he showed using the result by George Lowther, $$(*)\qquad P(|S|\ge1)\ge\frac{2a^2}{3(1+a)^2},$$ which actually holds for all possible values of $a$ (in $[0,1]$).

The main new ingredient is the following lower bound on $ES^4$:
$$(**)\qquad ES^4=3\Big(\sum_i a_i^2\Big)^2-2\sum_i a_i^4\ge3-2a^2, $$ which suggests that $|S|^2$ must be "large enough with a large enough probability" (as compared with $E|S|^2=1$). On the other hand, $$ ES^4=ES^4\,\mathrm{I}\{|S|<1\}+ES^4\,\mathrm{I}\{|S|\ge1\}\le1 +\sqrt{ES^8\,P(|S|\ge1)}, $$ by the Cauchy--Schwarz inequality; here $\mathrm{I}\{\cdot\}$ is the indicator function. Therefore and in view of $(**)$, $$(***)\qquad P(|S|\ge1)\ge\frac{(2-2a^2)^2}{ES^8}. $$ By the Whittle inequality Whittle, $ES^8\le EZ^8=105$, where $Z$ is a standard normal random variable. So, combining $(*)$ and $(***)$, one has $$ P(|S|\ge1)\ge\min_{a\in[0,1]}\Big(\frac{2a^2}{3(1+a)^2}\bigvee\frac{(2-2a^2)^2}{105}\Big)>0.032. $$

Addendum: As shown in Rademacher-lower, the lower bound $\frac{(2-2a^2)^2}{105}$ on $P(|S|\ge1)$ can be improved to $$\frac{2(1-a^2)^3}{(6+a^2)(5+2 a^2)}.$$ Thus, the lower bound $0.032$ gets improved to $0.043$.

Addendum 2: Instead of the Paley--Zygmund inequality, used by George Lowther to show that $P(|S|>u)\ge(1-u^2)^2/3$ for $u\in(0,1)$, one can use the Cantelli inequality to get $$P(|S|>u)=1-P(1-S^2\ge1-u^2)\ge\frac{(1-u^2)^2}{(1-u^2)^2+E(1-S^2)^2} \ge\frac{(1-u^2)^2}{(1-u^2)^2+2}, $$ since $E(1-S^2)^2=ES^4-1\le2$. Now proceeding as before, one has the lower bound $$\frac{a^2}{(1+a)^2+2a^2} $$ in place of $\frac{2a^2}{3(1+a)^2}$, which results in the improvement of the lower bound on $P(|S|\ge1)$ from $0.043$ to $0.04789$.

I would like indeed to improve the lower bound on $c$, to about $0.032$. I'll be using the same notations as in Anthony Quas's answer. As he showed using the result by George Lowther, $$(*)\qquad P(|S|\ge1)\ge\frac{2a^2}{3(1+a)^2},$$ which actually holds for all possible values of $a$ (in $[0,1]$).

The main new ingredient is the following lower bound on $ES^4$:
$$(**)\qquad ES^4=3\Big(\sum_i a_i^2\Big)^2-2\sum_i a_i^4\ge3-2a^2, $$ which suggests that $|S|^2$ must be "large enough with a large enough probability" (as compared with $E|S|^2=1$). On the other hand, $$ ES^4=ES^4\,\mathrm{I}\{|S|<1\}+ES^4\,\mathrm{I}\{|S|\ge1\}\le1 +\sqrt{ES^8\,P(|S|\ge1)}, $$ by the Cauchy--Schwarz inequality; here $\mathrm{I}\{\cdot\}$ is the indicator function. Therefore and in view of $(**)$, $$(***)\qquad P(|S|\ge1)\ge\frac{(2-2a^2)^2}{ES^8}. $$ By the Whittle inequality Whittle, $ES^8\le EZ^8=105$, where $Z$ is a standard normal random variable. So, combining $(*)$ and $(***)$, one has $$ P(|S|\ge1)\ge\min_{a\in[0,1]}\Big(\frac{2a^2}{3(1+a)^2}\bigvee\frac{(2-2a^2)^2}{105}\Big)>0.032. $$

Addendum: As shown in Rademacher-lower, the lower bound $\frac{(2-2a^2)^2}{105}$ on $P(|S|\ge1)$ can be improved to $$\frac{2(1-a^2)^3}{(6+a^2)(5+2 a^2)}.$$ Thus, the lower bound $0.032$ gets improved to $0.043$.

I would like indeed to improve the lower bound on $c$, to about $0.032$. I'll be using the same notations as in Anthony Quas's answer. As he showed using the result by George Lowther, $$(*)\qquad P(|S|\ge1)\ge\frac{2a^2}{3(1+a)^2},$$ which actually holds for all possible values of $a$ (in $[0,1]$).

The main new ingredient is the following lower bound on $ES^4$:
$$(**)\qquad ES^4=3\Big(\sum_i a_i^2\Big)^2-2\sum_i a_i^4\ge3-2a^2, $$ which suggests that $|S|^2$ must be "large enough with a large enough probability" (as compared with $E|S|^2=1$). On the other hand, $$ ES^4=ES^4\,\mathrm{I}\{|S|<1\}+ES^4\,\mathrm{I}\{|S|\ge1\}\le1 +\sqrt{ES^8\,P(|S|\ge1)}, $$ by the Cauchy--Schwarz inequality; here $\mathrm{I}\{\cdot\}$ is the indicator function. Therefore and in view of $(**)$, $$(***)\qquad P(|S|\ge1)\ge\frac{(2-2a^2)^2}{ES^8}. $$ By the Whittle inequality Whittle, $ES^8\le EZ^8=105$, where $Z$ is a standard normal random variable. So, combining $(*)$ and $(***)$, one has $$ P(|S|\ge1)\ge\min_{a\in[0,1]}\Big(\frac{2a^2}{3(1+a)^2}\bigvee\frac{(2-2a^2)^2}{105}\Big)>0.032. $$

Addendum: As shown in Rademacher-lower, the lower bound $\frac{(2-2a^2)^2}{105}$ on $P(|S|\ge1)$ can be improved to $$\frac{2(1-a^2)^3}{(6+a^2)(5+2 a^2)}.$$ Thus, the lower bound $0.032$ gets improved to $0.043$.

Addendum 2: Instead of the Paley--Zygmund inequality, used by George Lowther to show that $P(|S|>u)\ge(1-u^2)^2/3$ for $u\in(0,1)$, one can use the Cantelli inequality to get $$P(|S|>u)=1-P(1-S^2\ge1-u^2)\ge\frac{(1-u^2)^2}{(1-u^2)^2+E(1-S^2)^2} \ge\frac{(1-u^2)^2}{(1-u^2)^2+2}, $$ since $E(1-S^2)^2=ES^4-1\le2$. Now proceeding as before, one has the lower bound $$\frac{a^2}{(1+a)^2+2a^2} $$ in place of $\frac{2a^2}{3(1+a)^2}$, which results in the improvement of the lower bound on $P(|S|\ge1)$ from $0.043$ to $0.04789$.