I know that textbooks state that when conducting level-log regression, I should generally divide the estimated coefficient by Beta / 100 for a percentage point interpretation. For example, a 1 % increase in X leads to a percentage point change in Y. However, when we have a linear probability model, where y={0,1}, is it still necessary to divide the estimated log(X) coefficient by 100 for the same percentage point interpretation as the one stated above? My intuitive answer is no, it is no longer necessary in the linear probability model case, but I am not quite sure if I am correct or why I am correct. Thank you in advance for your clarification.
123094
If I have a level-log model, such as y= β0+β1log(x),  I know the interpretation is "If we increase x by one percent, we expect y to increase by (β1/100) units of y."  So, my questions are   If y is a percentage variable, what is the correct interpretation of the resulting regression? Is it still the same but has a unit of percentage instead of unit, so the interpretation becomes "If we increase x by one percent, we expect y to increase by (β1/100) percent of y"? Or we don’t have to divide β1 by 100, so the interpretation becomes "If we increase x by one percent, we expect y to increase by β1 percent of y"? Can I convert my variables (ex: interest rate and inflation rate) to logarithm if it has a unit of percentage? If it can be done, can I change my model into log-log model, although my variable y has a unit of percentage, so the model becomes  log(y)= β0+β1 log(x) ? What is the correct interpretation of this model?
30352
I'm wondering if it makes a difference in interpretation whether only the dependent, both the dependent and independent, or only the independent variables are log transformed.  Consider the case of  log(DV) = Intercept + B1*IV + Error    I can interpret the IV as the percent increase but how does this change when I have   log(DV) = Intercept + B1*log(IV) + Error   or when I have   DV = Intercept + B1*log(IV) + Error   ?
12261
As explained in , a linear model can be written in the form:  $$ y = \beta_1 x_{1} + \cdots + \beta_p x_{p} + \varepsilon_i,$$  where $y$ is the response variable and  $x_{i}$ is the $i^{th}$ explanatory variable.  Often with the goal of meeting test assumptions, one can transform the response variable. For example we apply the log function on each $y_i$. Transforming a response variable does NOT equate to doing a GLM.  A GLM can be written in the following form (from the )  $$ g(u) = \beta_1 x_{1} + \cdots + \beta_p x_{p} + \varepsilon_i,$$  where $u$ is just another symbol for $y$ as I understand from page 2 in the course handout. $g()$ is called the link function.  I don't really understand the difference between a GLM and LM with transformed variable from the slides in the course. Can you help me with that?
11222
Suppose   $$ (a,b) $$   is a $(1-\alpha)$ level confidence interval for a parameter $\theta$. Suppose $\eta$ is a monotone invertible transformation. Then, is   $$ \left (\eta(a), \eta(b) \right ) $$   a $(1-\alpha)$ level confidence interval for $\eta(\theta)$? Assume the parameter and the confidence interval endpoints are all real numbers.   The answer seems intuitively to be "Yes" for similar reasons to why you can transform random variables do things like, if $Y = g(X)$, then  $$ P( Y \leq y ) = P( g(X) \leq y ) = P(X \leq g^{-1}(y) ) $$   Might also be related to the continuous mapping theorem, as applied to MLEs.   This is not homework and is coming up in the context of whether or not I can nake a 95% for the log odds, then back-transform it and call it a 95% for the probability.   Thanks
40614
If $x=16384$ and $y=2$  $\ln(x)=9.704$ $\ln(y)=0.6931$  $\log(x)=4.2144$ $\log(y)=0.3010$  If we divide $\frac{\ln(x)}{\ln(y)}$ we get $14$ and same answer for $\frac{\log(x)}{\log(y)}$.  So can anyone tell me the concept behind this? Why does dividing $\frac{\ln(x)}{\ln(y)}$ give the same result as $\frac{\log(x)}{\log(y)}$?
79370
I have carried out a linear regression. The plot below shows the distribution of the response variable:    I believe the response variable is beta distributed, therefore virtually the exact opposite of normally distributed. However, when including all my predictor variables in the linear regression, the residuals turn out to be quite normally distributed, as shown in this plot:    Has my model satisfied the assumptions of linear regression? Might there be a better model to use?
50690
I am writing an algorithm where, given a model, I compute likelihoods for a list of datasets and then need to normalize (to probability) each one of the likelihood. So something like [0.00043, 0.00004, 0.00321] might be converted to may be like [0.2, 0.03, 0.77].   My problem is that the log likelihoods, I am working with, are quite small (for instance, in log space, the values are like -269647.432, -231444.981 etc). In my C++ code, when I try to add two of them (by taking their exponent) I get an answer of "Inf". I tried to add them in log-space , but again stumbled upon the same problem.  Can anybody share his/her expert opinion on this?
25999
Goal: create a plot to compare the empirical convergence rate of a numerical method with the one expected from the theory.  Accordingly, given a loglog-plot, I would like to add the line which has a prescribed slope (in this case, the convergence of the method) and the best intercept to fit the data (e.g., in a least-square sense).  Question: how to do a linear regression with fixed slope in pgfplots?  What I have done so far is the following: (1) compute the regression line (of the logarithm of the quantities), (2) save the intercept, (3) plot the line with the slope expected from the theory and this intercept.  This is a good approximation of what I am looking for, provided the data are in agreement with the theory.  Thanks  MdFB
27542
We usually get an estimate of $\beta$ in the logistic regression by finding the $MLE$ of the observed random samples of $X_1,X_2....,X_N$. Then we use Wald's test i.e. ${[\hat \beta / S.E.(\hat \beta)]}^2$ to test whether that variable is significant or not.   From what I have read, this Wald's test is based on two facts (or assumptions, I am not sure).    $\hat \beta$ follows a normal distribution Standard Deviation of this normal distribution is given by the inverse of Fisher's Matrix   Can someone explain the proofs behind these two facts/ (or assumptions). I have read this  but most of the intuition went over my head.
97783
I'm trying to get caught up in some notes, and the topic is pooled, cross-sectional data. If we have the regression: $$\log{wage_i}=\beta_0+\beta_1\text{married}_i+\delta_0\text{yr10}_i+\delta_1\text{married}_i\times \text{yr10}_i+\varepsilon_i$$ Where $\text{married}$ is a dummy variable which $=1$ if a person is married and $0$ if not, and $\text{yr10}$ is a dummy variable indicating whether the observation was recorded in the year 2010, as opposed to being recorded in the year 2000. There's a blank spot in the notes that I didn't get, and it asks: How do we interpret $\delta_1$?  I think it's correct to say that $\delta_1$ measures how the return on whether or not someones married has changed in the past $10$ years. Is that correct? Or am I wrong? I read a little bit in my textbook and I think that's what it's saying.
29978
This is somewhat of a beginner's question, but how does one interpret an exp(B) result of 6.012 in a multinomial logistic regression model?  1) is it 6.012-1.0 = 5.012 = 5012% increase in risk?  or  2) 6.012/(1+6.012) = 0.857 = 85.7% increase in risk?  In case both alternatives are incorrect, can someone please mention the correct way?  Ive searched many resources on the internet and I get to these two alternatives, and I am not entirely sure which one is correct.
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In practice, using a standard T-test to check the significance of a linear regression coefficient is common practice. The mechanics of the calculation make sense to me.  Why is it that the T-distribution can be used to model the standard test statistic used in linear regression hypothesis testing? Standard test statistic I am referring to here:   $$ T_{0} = \frac{\widehat{\beta} - \beta_{0}}{SE(\widehat{\beta})} $$
96236
I start with my OLS regression: $$ y = \beta _0 + \beta_1x_1+\beta_2 D + \varepsilon  $$ where D is a dummy variable, the estimates become different from zero with a low p-value. I then preform a Ramsey RESET test and find that i have some misspesification of the equation, i thus include squared x: $$ y = \beta _0 + \beta_1x_1+\beta_2x_1^2+\beta_3 D + \varepsilon  $$   What does the squared term explain? (Non-linear increase in Y?)  By doing this my D estimate does not vary from zero any more, with a high p-value. How do i interpret the squared term in my equation (in general)?   Edit: Improving question.
122317
I am currently reading a  concerning voting location and voting preference in the 2000 and 2004 election. In it, there is a chart which displays logistic regression coefficients.  From courses years back and a little , I understand logistic regression to be a way of describing the relationship between multiple independent variables and a binary response variable. What I'm confused about is, given the table below, because the South has a logistic regression coefficient of .903, does that mean that 90.3% of Southerners vote republican? Because of the logistical nature of the metric, that this direct correlation does not exist. Instead, I assume that you can only say that the south, with .903, votes Republican more than the Mountains/plains, with the regression of .506.  Given the latter to be the case, how do I know what is significant and what is not and is it possible to extrapolate a percentage of republican votes given this logistic regression coefficient.  As a side note, please edit my post if anything is stated incorrectly
10775
I'm studying statistics and often come across formulae containing the log and I'm always confused if I should interpret that as the standard meaning of log, i.e. base 10, or if in statistics the symbol log  is generally assumed to be the natural log ln.  In particular I'm studying the  as an example, but my question is more of general one.
23544
I have a simple linear OLS regression $Y_i = \alpha+ \beta_1 X_{1i} + \beta_2 X_{2i} + e_i$ where $e_i \sim N(0,\sigma)$. I have estimated the regression from the data and obtained estimates for my coefficients as well as the corresponding covariance matrix. Assume my dataset is large (>500)  Now I would like to construct the 95% confidence interval for $Z = a_1 \beta_1 + a_2 \beta_2$. My first thought was that I could assume that Z follows a normal distribution because my regression coefficients are approximately normal and a linear combination of normal variables is also a normal variable. I do find some sources where they treat $Z$ as a normal random variable.  But now I am in doubt because I found that a linear combination of normal random variables is only normal if they are independent. So is it still save to assume $Z$ is random?
73007
I know I mess up different things. However I want to get better understanding what is motivation behind logistic regression.  $p(y_i=1|x_i)=(1+e^{-x_iw})^{-1}$  and according to OLS  $S(w)=\sum_{i}^{}(y_i-\hat{y_i})^2$,where $y \in \{1,0\}$.  What if $\hat{y_i}=p(y_i=1|x_i)$ and estimate $w$ by finding minimum to $S(w)$.  Why this is incorrect approach?