Probstat/midterm problem 9 solution - ประวัติรุ่นแก้ไข

Jittat: /* Variance */

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Variance

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Variance

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Variance

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Variance

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Variance

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Variance

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Variance

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Variance

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Variance

@@ แถว 96: / แถว 96: @@
 </center>
-which is
+which is ...
 == Solution 2 ==

@@ แถว 93: / แถว 93: @@
 <center>
-<math>Var(X) = E[X^2] - E[X]^2 = \left(\frac{n^2}{2n-1}\right) + \left( \frac{n(n-1)}{2}\cdot \frac{n}{2n-1}\cdot\frac{n-1}{2n-3} \right) - (\left(\frac{n^2}{2n-1}\right))^2,</math>
+<math>Var(X) = E[X^2] - E[X]^2 = \left(\frac{n^2}{2n-1}\right) + \left( \frac{n(n-1)}{2}\cdot \frac{n}{2n-1}\cdot\frac{n-1}{2n-3} \right) - \left(\frac{n^2}{2n-1}\right)^2,</math>
 </center>

@@ แถว 89: / แถว 89: @@
 </math>
 </center>
 == Solution 2 ==

@@ แถว 84: / แถว 84: @@
 <math>E[X^2]
+=
-\sum_{i=1}^n \frac{n}{2n-1} + \sum_{i=1}^n\sum_{ j\neq i } \frac{n}{2n-1}\cdot\frac{n-1}{2n-3}
+\left(\sum_{i=1}^n \frac{n}{2n-1}\right) + \left(\sum_{i=1}^n\sum_{ j\neq i } \frac{n}{2n-1}\cdot\frac{n-1}{2n-3}\right)
+=
-\frac{n^2}{2n-1} + \frac{n(n-1)}{2}\cdot \frac{n}{2n-1}\cdot\frac{n-1}{2n-3}
+\left(\frac{n^2}{2n-1}\right) + \left( \frac{n(n-1)}{2}\cdot \frac{n}{2n-1}\cdot\frac{n-1}{2n-3} \right)
 </math>
 </center>

@@ แถว 82: / แถว 82: @@
 <center>
-<math>E[X^2] =
+<math>E[X^2]
-\sum_{i=1}^n \frac{n}{2n-1} + \sum_{i=1}^n\sum_{ j\neq i } \frac{n}{2n-1}\cdot\frac{n-1}{2n-3}.
 </math>
 </center>

@@ แถว 77: / แถว 77: @@
 E\left[\sum_{i=1}^n X_i^2 + \sum_{i=1}^n\sum_{ j\neq i } X_iX_j\right] = \sum_{i=1}^n E\left[X_i^2\right] + \sum_{i=1}^n\sum_{ j\neq i } E\left[X_iX_j\right].
 </math>
 </center>

@@ แถว 74: / แถว 74: @@
 <center>
 <math>
-E[X^2] = E\eft[\left(\sum_{i=1}^n X_i\right)^2\right] =
+E[X^2] = E\left[\left(\sum_{i=1}^n X_i\right)^2\right] =
 E\left[\sum_{i=1}^n X_i^2 + \sum_{i=1}^n\sum_{ j\neq i } X_iX_j\right] = \sum_{i=1}^n E\left[X_i^2\right] + \sum_{i=1}^n\sum_{ j\neq i } E\left[X_iX_j\right].
 </math>

@@ แถว 60: / แถว 60: @@
 </center>
-and from the previous solution, we conclude that <math>E[X_i^2] = \frac{n}{2n-1}.</math>.
+and from the previous solution, we conclude that <math>E[X_i^2] = \frac{n}{2n-1}.</math>
 Let's consider the second term, <math>E[X_iX_j]</math>.  Again note that the only way <math>X_iX_j</math> is non zero is when both <math>X_i</math> and <math>X_j</math> are one.  You can check that this occurs with probability
@@ แถว 66: / แถว 66: @@
 <center>
 <math>\frac{n}{2n-1}\cdot\frac{n-1}{2n-3}</math>.
 </center>