Updating typo in statistics slides

This commit is contained in:
mhjensen
2018-08-24 06:45:18 +02:00
parent 17f4c9ec81
commit d0625170e7
9 changed files with 10 additions and 10 deletions
@@ -411,8 +411,8 @@ while \( P(x) \) is the cumulative probability.
<tr><td align="left"> Domain </td> <td align="center"> \( \left\{x_1, x_2, x_3, \dots, x_N\right\} \) </td> <td align="center"> \( [a,b] \) </td> </tr>
<tr><td align="left"> Probability </td> <td align="center"> \( p(x_i) \) </td> <td align="center"> \( p(x)dx \) </td> </tr>
<tr><td align="left"> Cumulative </td> <td align="center"> \( P_i=\sum_{l=1}^ip(x_l) \) </td> <td align="center"> \( P(x)=\int_a^xp(t)dt \) </td> </tr>
<tr><td align="left"> Positivity </td> <td align="center"> $ 0 \le p(x_i) \le 1$ </td> <td align="center"> $ p(x) \ge 0$ </td> </tr>
<tr><td align="left"> Positivity </td> <td align="center"> $ 0 \le P_i \le 1$ </td> <td align="center"> $ 0 \le P(x) \le 1$ </td> </tr>
<tr><td align="left"> Positivity </td> <td align="center"> \( 0 \le p(x_i) \le 1 \) </td> <td align="center"> \( p(x) \ge 0 \) </td> </tr>
<tr><td align="left"> Positivity </td> <td align="center"> \( 0 \le P_i \le 1 \) </td> <td align="center"> \( 0 \le P(x) \le 1 \) </td> </tr>
<tr><td align="left"> Monotonic </td> <td align="center"> \( P_i \ge P_j \) if \( x_i \ge x_j \) </td> <td align="center"> \( P(x_i) \ge P(x_j) \) if \( x_i \ge x_j \) </td> </tr>
<tr><td align="left"> Normalization </td> <td align="center"> \( P_N=1 \) </td> <td align="center"> \( P(b)=1 \) </td> </tr>
</tbody>
@@ -616,8 +616,8 @@ while \( P(x) \) is the cumulative probability.
<tr><td align="left"> Domain </td> <td align="center"> \( \left\{x_1, x_2, x_3, \dots, x_N\right\} \) </td> <td align="center"> \( [a,b] \) </td> </tr>
<tr><td align="left"> Probability </td> <td align="center"> \( p(x_i) \) </td> <td align="center"> \( p(x)dx \) </td> </tr>
<tr><td align="left"> Cumulative </td> <td align="center"> \( P_i=\sum_{l=1}^ip(x_l) \) </td> <td align="center"> \( P(x)=\int_a^xp(t)dt \) </td> </tr>
<tr><td align="left"> Positivity </td> <td align="center"> $ 0 \le p(x_i) \le 1$ </td> <td align="center"> $ p(x) \ge 0$ </td> </tr>
<tr><td align="left"> Positivity </td> <td align="center"> $ 0 \le P_i \le 1$ </td> <td align="center"> $ 0 \le P(x) \le 1$ </td> </tr>
<tr><td align="left"> Positivity </td> <td align="center"> \( 0 \le p(x_i) \le 1 \) </td> <td align="center"> \( p(x) \ge 0 \) </td> </tr>
<tr><td align="left"> Positivity </td> <td align="center"> \( 0 \le P_i \le 1 \) </td> <td align="center"> \( 0 \le P(x) \le 1 \) </td> </tr>
<tr><td align="left"> Monotonic </td> <td align="center"> \( P_i \ge P_j \) if \( x_i \ge x_j \) </td> <td align="center"> \( P(x_i) \ge P(x_j) \) if \( x_i \ge x_j \) </td> </tr>
<tr><td align="left"> Normalization </td> <td align="center"> \( P_N=1 \) </td> <td align="center"> \( P(b)=1 \) </td> </tr>
</tbody>
@@ -748,8 +748,8 @@ while \( P(x) \) is the cumulative probability.
<tr><td align="left"> Domain </td> <td align="center"> \( \left\{x_1, x_2, x_3, \dots, x_N\right\} \) </td> <td align="center"> \( [a,b] \) </td> </tr>
<tr><td align="left"> Probability </td> <td align="center"> \( p(x_i) \) </td> <td align="center"> \( p(x)dx \) </td> </tr>
<tr><td align="left"> Cumulative </td> <td align="center"> \( P_i=\sum_{l=1}^ip(x_l) \) </td> <td align="center"> \( P(x)=\int_a^xp(t)dt \) </td> </tr>
<tr><td align="left"> Positivity </td> <td align="center"> $ 0 \le p(x_i) \le 1$ </td> <td align="center"> $ p(x) \ge 0$ </td> </tr>
<tr><td align="left"> Positivity </td> <td align="center"> $ 0 \le P_i \le 1$ </td> <td align="center"> $ 0 \le P(x) \le 1$ </td> </tr>
<tr><td align="left"> Positivity </td> <td align="center"> \( 0 \le p(x_i) \le 1 \) </td> <td align="center"> \( p(x) \ge 0 \) </td> </tr>
<tr><td align="left"> Positivity </td> <td align="center"> \( 0 \le P_i \le 1 \) </td> <td align="center"> \( 0 \le P(x) \le 1 \) </td> </tr>
<tr><td align="left"> Monotonic </td> <td align="center"> \( P_i \ge P_j \) if \( x_i \ge x_j \) </td> <td align="center"> \( P(x_i) \ge P(x_j) \) if \( x_i \ge x_j \) </td> </tr>
<tr><td align="left"> Normalization </td> <td align="center"> \( P_N=1 \) </td> <td align="center"> \( P(b)=1 \) </td> </tr>
</tbody>
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@@ -753,8 +753,8 @@ while \( P(x) \) is the cumulative probability.
<tr><td align="left"> Domain </td> <td align="center"> \( \left\{x_1, x_2, x_3, \dots, x_N\right\} \) </td> <td align="center"> \( [a,b] \) </td> </tr>
<tr><td align="left"> Probability </td> <td align="center"> \( p(x_i) \) </td> <td align="center"> \( p(x)dx \) </td> </tr>
<tr><td align="left"> Cumulative </td> <td align="center"> \( P_i=\sum_{l=1}^ip(x_l) \) </td> <td align="center"> \( P(x)=\int_a^xp(t)dt \) </td> </tr>
<tr><td align="left"> Positivity </td> <td align="center"> $ 0 \le p(x_i) \le 1$ </td> <td align="center"> $ p(x) \ge 0$ </td> </tr>
<tr><td align="left"> Positivity </td> <td align="center"> $ 0 \le P_i \le 1$ </td> <td align="center"> $ 0 \le P(x) \le 1$ </td> </tr>
<tr><td align="left"> Positivity </td> <td align="center"> \( 0 \le p(x_i) \le 1 \) </td> <td align="center"> \( p(x) \ge 0 \) </td> </tr>
<tr><td align="left"> Positivity </td> <td align="center"> \( 0 \le P_i \le 1 \) </td> <td align="center"> \( 0 \le P(x) \le 1 \) </td> </tr>
<tr><td align="left"> Monotonic </td> <td align="center"> \( P_i \ge P_j \) if \( x_i \ge x_j \) </td> <td align="center"> \( P(x_i) \ge P(x_j) \) if \( x_i \ge x_j \) </td> </tr>
<tr><td align="left"> Normalization </td> <td align="center"> \( P_N=1 \) </td> <td align="center"> \( P(b)=1 \) </td> </tr>
</tbody>
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@@ -332,8 +332,8 @@ while $P(x)$ is the cumulative probability.
| Domain | $\left\{x_1, x_2, x_3, \dots, x_N\right\}$ | $[a,b]$ |
| Probability | $p(x_i)$ | $p(x)dx$ |
| Cumulative | $P_i=\sum_{l=1}^ip(x_l)$ | $P(x)=\int_a^xp(t)dt$ |
| Positivity | $ 0 \le p(x_i) \le 1$ | $ p(x) \ge 0$ |
| Positivity | $ 0 \le P_i \le 1$ | $ 0 \le P(x) \le 1$ |
| Positivity | $0 \le p(x_i) \le 1$ | $p(x) \ge 0$ |
| Positivity | $0 \le P_i \le 1$ | $0 \le P(x) \le 1$ |
| Monotonic | $P_i \ge P_j$ if $x_i \ge x_j$ | $P(x_i) \ge P(x_j)$ if $x_i \ge x_j$ |
| Normalization | $P_N=1$ | $P(b)=1$ |
|--------------------------------------------------------------------------------------------------------------------------------------|