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Iosif Pinelis
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Note that $$I_n^*\le E(X_1+\cdots+X_n)=n/2.$$ This upper bound $n/2$ on $I_n^*$ is attained if $n=2m$ is even and $X_{2i}=1-X_{2i-1}$ for $i=1,\dots,m$.

So, $$I_n^*=n/2$$ for even $n$.

It also follows that $$(n-1)/2=I_{n-1}^*\le I_n^*\le n/2$$ for odd $n$.

So, $$I_n^*\sim n/2$$ for $\Bbb N\ni n\to\infty$.


Conjecture: $I_n^*=n/2$ for all natural $n\ge2$.

Remark 1: Obviously, $I_1^*=0$. In view of the above, it is enough to show that $I_n^*=n/2$ for all odd $n\ge3$. Moreover, in view of the above cancellations, it is enough to show that $I_3^*=3/2$.

Note that $$I_n^*\le E(X_1+\cdots+X_n)=n/2.$$ This upper bound $n/2$ on $I_n^*$ is attained if $n=2m$ is even and $X_{2i}=1-X_{2i-1}$ for $i=1,\dots,m$.

So, $$I_n^*=n/2$$ for even $n$.

It also follows that $$(n-1)/2=I_{n-1}^*\le I_n^*\le n/2$$ for odd $n$.

So, $$I_n^*\sim n/2$$ for $\Bbb N\ni n\to\infty$.

Note that $$I_n^*\le E(X_1+\cdots+X_n)=n/2.$$ This upper bound $n/2$ on $I_n^*$ is attained if $n=2m$ is even and $X_{2i}=1-X_{2i-1}$ for $i=1,\dots,m$.

So, $$I_n^*=n/2$$ for even $n$.

It also follows that $$(n-1)/2=I_{n-1}^*\le I_n^*\le n/2$$ for odd $n$.

So, $$I_n^*\sim n/2$$ for $\Bbb N\ni n\to\infty$.


Conjecture: $I_n^*=n/2$ for all natural $n\ge2$.

Remark 1: Obviously, $I_1^*=0$. In view of the above, it is enough to show that $I_n^*=n/2$ for all odd $n\ge3$. Moreover, in view of the above cancellations, it is enough to show that $I_3^*=3/2$.

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Iosif Pinelis
  • 128k
  • 8
  • 107
  • 229

Note that $$I_n^*\le E(X_1+\cdots+X_n)=n/2.$$ This upper bound $n/2$ on $I_n^*$ is attained if $n=2m$ is even and $X_{2i}=1-X_{2i-1}$ for $i=1,\dots,m$.

So, $$I_n^*=n/2$$ for even $n$.

It also follows that $$(n-1)/2=I_{n-1}^*\le I_n^*\le n/2$$ for odd $n$.

So, $$I_n^*\sim n/2$$ for $\Bbb N\ni n\to\infty$.

Note that $$I_n^*\le E(X_1+\cdots+X_n)=n/2.$$ This upper bound on $I_n^*$ is attained if $n=2m$ is even and $X_{2i}=1-X_{2i-1}$ for $i=1,\dots,m$.

So, $$I_n^*=n/2$$ for even $n$.

It also follows that $$(n-1)/2=I_{n-1}^*\le I_n^*\le n/2$$ for odd $n$.

So, $$I_n^*\sim n/2$$ for $\Bbb N\ni n\to\infty$.

Note that $$I_n^*\le E(X_1+\cdots+X_n)=n/2.$$ This upper bound $n/2$ on $I_n^*$ is attained if $n=2m$ is even and $X_{2i}=1-X_{2i-1}$ for $i=1,\dots,m$.

So, $$I_n^*=n/2$$ for even $n$.

It also follows that $$(n-1)/2=I_{n-1}^*\le I_n^*\le n/2$$ for odd $n$.

So, $$I_n^*\sim n/2$$ for $\Bbb N\ni n\to\infty$.

Source Link
Iosif Pinelis
  • 128k
  • 8
  • 107
  • 229

Note that $$I_n^*\le E(X_1+\cdots+X_n)=n/2.$$ This upper bound on $I_n^*$ is attained if $n=2m$ is even and $X_{2i}=1-X_{2i-1}$ for $i=1,\dots,m$.

So, $$I_n^*=n/2$$ for even $n$.

It also follows that $$(n-1)/2=I_{n-1}^*\le I_n^*\le n/2$$ for odd $n$.

So, $$I_n^*\sim n/2$$ for $\Bbb N\ni n\to\infty$.