minimum mean square error of $X_{1}$ given $S_{n}=\sum_{i=1}^{n}X_{i}$ when $p=0.5$ and $X_{1},X_{2},\ldots$ are bernoulli iid

Question

I need help with this one: Let $X_{1},X_{2},\ldots$ be bernoulli i.i.d random variables with $$\forall i:\mathbb{P}\left\{ X_{i}=1\right\} =\mathbb{P}\left\{ X_{i}=0\right\} =\frac{1}{2}$$ define $S_{n}=\sum_{i=1}^{n}X_{i}$ the qustion was to calculate the optimal estimator of $X_{1}$ given $S_{n}$ and calculate it's error (it's the minimum mean square error). Iv'e calculated already the optimal estimator of $X_{1}$ given $S_{n}$ ,I know it's $$\hat{X}_{S_{n}}^{\text{opt}}=\mathbb{E}\left[X_{1}\mid S_{n}=s\right]=\frac{S_{n}}{n}$$ and I calculate the minimum mean square error: It's $$\epsilon^{2}\left(\hat{X}_{S_{n}}^{\text{opt}}\right)=\mathbb{E}\left[X_{1}^{2}\right]-\mathbb{E}\left[\left(\hat{X}_{S_{n}}^{\text{opt}}\right)^{2}\right]$$ when $$\mathbb{E}\left[\hat{X}_{\text{opt}}^{2}\right] =\mathbb{E}\left[\left(\frac{S_{n}}{n}\right)^{2}\right] =\frac{1}{n^{2}}\cdot\mathbb{E}\left[\left(S_{n}\right)^{2}\right] =\frac{1}{n^{2}}\cdot\mathbb{E}\left[\left(\sum_{i=1}^{n}X_{i}\right)^{2}\right] =\frac{1}{n^{2}}\cdot\mathbb{E}\left[\sum_{i=1}^{n}\sum_{j=1}^{n}X_{j}X_{i}\right] =\frac{1}{n^{2}}\cdot\mathbb{E}\left[\sum_{i=1}^{n}\sum_{j\in\left\{ 0,\ldots,n\right\} \backslash\left\{ i\right\} }X_{j}X_{i}+\sum_{i=1}^{n}\left(X_{i}^{2}\right)\right] =\ldots =\frac{1}{n^{2}}\cdot\left[\sum_{i=1}^{n}\sum_{j\in\left\{ 0,\ldots,n\right\} \backslash\left\{ i\right\} }\frac{1}{2}\cdot\frac{1}{2}+\frac{1}{2}n\right] =\frac{1}{n^{2}}\cdot\left[\sum_{i=1}^{n}\sum_{j\in\left\{ 0,\ldots,n\right\} \backslash\left\{ i\right\} }\frac{1}{2}\cdot\frac{1}{2}+\frac{1}{2}n\right] =\frac{1}{n^{2}}\cdot\left[n\left(n-1\right)\frac{1}{4}+\frac{1}{2}n\right] =\left(\frac{n-1}{n}\right)\frac{1}{4}+\frac{1}{2n} =\left(1-\frac{1}{n}\right)\frac{1}{4}+\frac{1}{2n} =\frac{1}{4}-\frac{1}{n}\frac{1}{4}+\frac{1}{2n} =\frac{1}{4}-\frac{1}{n}\frac{1}{4}+\frac{2}{4n} =\frac{1}{4}+\frac{1}{4n}$$ and $$\mathbb{E}\left[X_{1}^{2}\right]=\mathbb{P}\left\{ X_{1}=1\right\} \cdot\left(1\right)^{2}+\mathbb{P}\left\{ X_{1}=0\right\} \cdot\left(0\right)^{2}=\frac{1}{2}\cdot1^{2}+\frac{1}{2}\cdot0^{2}=\frac{1}{2}$$ so i plug it in to the error function and get: $$\epsilon^{2}\left(\hat{X}_{S_{n}}^{\text{opt}}\right)=\mathbb{E}\left[X_{1}^{2}\right]-\mathbb{E}\left[\left(\hat{X}_{S_{n}}^{\text{opt}}\right)^{2}\right]=\frac{1}{2}-\left(\frac{1}{4}+\frac{1}{4n}\right)=\frac{1}{4}-\frac{1}{4n}$$ but my intonation doesn't fit here:

1. if $n=1$ we know $X_{1}$ for sure : It's $S_{n}$ and we get no error( that's ok with my error term)
2. but if $n\rightarrow\infty$ I got that $$\epsilon^{2}\left(\hat{X}_{S_{n}}^{\text{opt}}\right)\approx\frac{1}{4}$$ .that's densest make sense becuse $\frac{S_{n}}{n}\xrightarrow{n\rightarrow\infty}\frac{1}{2}$ and in that case $\hat{X}_{S_{n}}^{\text{opt}}\approx\frac{1}{2}$ and we know that $X_{1}\in\left\{ 0,1\right\} $ so the error is also need to be $\epsilon^{2}\approx\frac{1}{2}$ in that case...

what I'm missing here? my intonation is right?I think my error calculation is wrong.

Thanks!

Answer 1

I think your intuition and error calculation are correct. What you have computed is the Mean "Squared" error $\epsilon^{2}\left(\hat{X}_{S_{n}}^{\text{opt}}\right)\approx\frac{1}{4}$ as $ n \rightarrow \infty$. Hence the error $\epsilon \approx \frac{1}{2}$ as $ n \rightarrow \infty$.