4
Say we have a model like this:
$$\log\hat y=f(x)=\beta_0+\beta_1 x$$
Since $\exp$ is the inverse function of $\log$, we can do this:
$$f(x)=\exp(\beta_0+\beta_1 x)$$
Now, what happens when $x$ grows by 1? $f(x)$ multiplies by $\exp(\beta_1)$:
$$\begin{align}
f(x+1)&=\exp[\beta_0+\beta_1(x+1)]\\
&=\exp(\beta_0+\beta_1 x)\cdot\exp(\beta_1)\\
&=f(x)\...
2
After the edits, I think the only remaining source of confusion is that is the prediction on the log scale, so instead we can write
$$
\begin{align}
\ln(\hat{b}) &= -4.40680302+0.00149165363 t \\
\hat{b}&=\exp(-4.40680302+0.00149165363t) \\
&= \exp(-4.40680302)\exp(0.00149165363t) \\
&= \hat{A}\exp(\hat{\beta} t)
\end{align}
$$
As for your ...
2
For simplicity, let's say you have a (binary) logistic regression model where you regress the binary outcome variable $Y$ on the predictors $X_1$ and $X_2$. $Y$ takes the value 1 for the event of interest (e.g., student is admitted into the program) and 0 otherwise (e.g., student is not admitted into the program).
The (binary) logistic regression model is ...
2
It is easy to misinterpret the coefficients in a regression involving interactions. Your regression results agree with the plot.
In standard treatment coding (the default in R), an intercept represents the outcome value when all continuous predictors are at 0 and categorical predictors are at their reference levels. The individual coefficients for non-...
2
In regression analysis, if you want to get asymptotic consistency results you need to impose some limiting conditions on the explanatory variables. The regression model itself makes no assumptions about the form of the sequence of explanatory variables, so these limiting conditions are conditions that go beyond the model assumptions for regression analysis. ...
2
The variance of a sum of random variables only equals the sum of their variances if the random variables are uncorrelated. If they are correlated a covariance term needs to be included:
$ var(X + Y) = var(X) + var(Y) + 2 cov(X, Y) $
where $X$ and $Y$ are random variables. This follows from the fact that the covariance is linear in each of its arguments ($\...
1
$Y=\beta_0 x^{\beta_1} e$ is not a linear model, it's a power model that can be linearized to a linear model, $\ln Y = \ln \beta_0+\beta_1 \ln x +\ln e$.
Doing that, and estimating it with least-squares, entail an assumption on the original model: $\ln e \sim \mathcal N (0, \sigma^2)$.
In other words, you assume that the $e$ terms are lognormal-distributed.
1
No, this is surely incorrect. Abstracting from all sorts of issues when forming causal statements (e.g. endogeneity) in the linear regression model (as mentioned in the comment correlation is not causation), LASSO, like any other penalized estimator induces bias due to the penalty term that is applied to regression coefficients. A bit more formally, suppose
$...
1
You might be best off using the bootstrap to validate your modeling process.
In outline, you repeat the entire modeling process--including the determination of optimal $\alpha$ and $\lambda$ values by cross validation--on multiple bootstrap samples of your data, then evaluate each model's performance on the entire data set. That mimics the process of ...
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