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2 changes: 1 addition & 1 deletion ch3/applied.Rmd
Original file line number Diff line number Diff line change
Expand Up @@ -609,7 +609,7 @@ x = c(coefficients(lm.zn)[2],
y = coefficients(lm.all)[2:14]
plot(x, y)
```
Coefficient for nox is approximately -10 in univariate model and 31 in
Coefficient for nox is approximately 31 in univariate model and -10 in
multiple regression model.

15d.
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1,233 changes: 654 additions & 579 deletions ch3/applied.html

Large diffs are not rendered by default.

7 changes: 2 additions & 5 deletions ch4/2.Rmd
Original file line number Diff line number Diff line change
Expand Up @@ -33,7 +33,7 @@ $$
\geq
\pi_i \exp (x \frac {\mu_i} {\sigma^2} - \frac {\mu_i^2} {2 \sigma^2})
$$
Multipy this inequality by the positive constant
Multiply this inequality by the positive constant
$$
c = \frac {
\frac {1} {\sqrt{2 \pi} \sigma}
Expand All @@ -47,7 +47,6 @@ c = \frac {
$$
and we have that
$$

\frac {\pi_k
\frac {1} {\sqrt{2 \pi} \sigma}
\exp(- \frac {1} {2 \sigma^2} (x - \mu_k)^2)
Expand All @@ -57,9 +56,7 @@ $$
\frac {1} {\sqrt{2 \pi} \sigma}
\exp(- \frac {1} {2 \sigma^2} (x - \mu_l)^2)
}}

\geq

\frac {\pi_i
\frac {1} {\sqrt{2 \pi} \sigma}
\exp(- \frac {1} {2 \sigma^2} (x - \mu_i)^2)
Expand All @@ -70,4 +67,4 @@ $$
\exp(- \frac {1} {2 \sigma^2} (x - \mu_l)^2)
}}
$$
or equivalently, $f_k(x) \geq f_i(x)$. Reversing these steps also holds, so we have that maximizing $\delta_k$ is equivalent to maximizing $p_k$.
or equivalently, $f_k(x) \geq f_i(x)$. Reversing these steps also holds, so we have that maximizing $\delta_k$ is equivalent to maximizing $p_k$.
272 changes: 128 additions & 144 deletions ch4/2.html

Large diffs are not rendered by default.

4 changes: 1 addition & 3 deletions ch4/2.md
Original file line number Diff line number Diff line change
Expand Up @@ -33,7 +33,7 @@ $$
\geq
\pi_i \exp (x \frac {\mu_i} {\sigma^2} - \frac {\mu_i^2} {2 \sigma^2})
$$
Multipy this inequality by the positive constant
Multiply this inequality by the positive constant
$$
c = \frac {
\frac {1} {\sqrt{2 \pi} \sigma}
Expand All @@ -57,9 +57,7 @@ $$
\frac {1} {\sqrt{2 \pi} \sigma}
\exp(- \frac {1} {2 \sigma^2} (x - \mu_l)^2)
}}

\geq

\frac {\pi_i
\frac {1} {\sqrt{2 \pi} \sigma}
\exp(- \frac {1} {2 \sigma^2} (x - \mu_i)^2)
Expand Down