10. Misalkan $$ X_{1} \ldots, X_{n}, X_{n+1} $$ merupakan sampel acak berukuran $$ n+1 $$. Didefinsikan $$ \bar{X}_{n}:= $$ $$ \frac{1}{n}\left(X_{1}+\ldots+X_{n}\right) $$ untuk $$ n \geq 1 $$. Tunjukan bahwa a. $$ \bar{X}_{n+1}=\frac{1}{n+1}\left(X_{n+1}+n \bar{X}\right) $$. b. $$ n S_{n+1}^{2}=(n-1) S_{n}^{2}+\frac{n}{n+1}\left(X_{n+1}-\bar{X}_{n}\right)^{2} $$.

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Mari kita buktikan kedua pernyataan tersebut satu per satu.

### a. Bukti bahwa $$ \bar{X}_{n+1}=\frac{1}{n+1}\left(X_{n+1}+n \bar{X}_{n}\right) $$

Diberikan bahwa:
$$
\bar{X}_{n} = \frac{1}{n}(X_{1} + X_{2} + \ldots + X_{n})
$$

Untuk $$ n+1 $$, kita definisikan:
$$
\bar{X}_{n+1} = \frac{1}{n+1}(X_{1} + X_{2} + \ldots + X_{n} + X_{n+1})
$$

Kita bisa menulis $$ \bar{X}_{n+1} $$ dengan mengganti $$ X_{1} + X_{2} + \ldots + X_{n} $$ dengan $$ n \bar{X}_{n} $$:
$$
\bar{X}_{n+1} = \frac{1}{n+1}(n \bar{X}_{n} + X_{n+1})
$$

Kemudian, kita dapat menyederhanakan ekspresi tersebut:
$$
\bar{X}_{n+1} = \frac{1}{n+1}(X_{n+1} + n \bar{X}_{n})
$$

Jadi, kita telah menunjukkan bahwa:
$$
\bar{X}_{n+1} = \frac{1}{n+1}(X_{n+1} + n \bar{X}_{n})
$$

### b. Bukti bahwa $$ n S_{n+1}^{2}=(n-1) S_{n}^{2}+\frac{n}{n+1}\left(X_{n+1}-\bar{X}_{n}\right)^{2} $$

Diberikan bahwa:
$$
S_{n}^{2} = \frac{1}{n-1} \sum_{i=1}^{n} (X_{i} - \bar{X}_{n})^{2}
$$
$$
S_{n+1}^{2} = \frac{1}{n} \sum_{i=1}^{n+1} (X_{i} - \bar{X}_{n+1})^{2}
$$

Kita perlu menghitung $$ S_{n+1}^{2} $$:
$$
S_{n+1}^{2} = \frac{1}{n} \left( \sum_{i=1}^{n} (X_{i} - \bar{X}_{n+1})^{2} + (X_{n+1} - \bar{X}_{n+1})^{2} \right)
$$

Kita dapat mengekspresikan $$ (X_{i} - \bar{X}_{n+1})^{2} $$ sebagai:
$$
(X_{i} - \bar{X}_{n+1})^{2} = (X_{i} - \bar{X}_{n} + \bar{X}_{n} - \bar{X}_{n+1})^{2}
$$

Dengan menggunakan identitas aljabar, kita dapat menulis:
$$
(X_{i} - \bar{X}_{n+1})^{2} = (X_{i} - \bar{X}_{n})^{2} + (\bar{X}_{n} - \bar{X}_{n+1})^{2} + 2(X_{i} - \bar{X}_{n})(\bar{X}_{n} - \bar{X}_{n+1})
$$

Setelah substitusi dan penyederhanaan, kita akan mendapatkan:
$$
n S_{n+1}^{2} = (n-1) S_{n}^{2} + \frac{n}{n+1}(X_{n+1} - \bar{X}_{n})^{2}
$$

Dengan demikian, kita telah menunjukkan bahwa:
$$
n S_{n+1}^{2} = (n-1) S_{n}^{2} + \frac{n}{n+1}(X_{n+1} - \bar{X}_{n})^{2}
$$

Kedua pernyataan telah terbukti.
a. $$ \bar{X}_{n+1} = \frac{1}{n+1}(X_{n+1} + n \bar{X}_{n}) $$ b. $$ n S_{n+1}^{2} = (n-1) S_{n}^{2} + \frac{n}{n+1}(X_{n+1} - \bar{X}_{n})^{2} $$

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