# Tag Info

1

The equation for the elasticity is: $e=\frac{dy}{dx}\frac{x}{y}$ In your case, we have $ln(y)=5-0.1x$, Now take the derivative of both sides and multiply by x: $\frac{dy}{dx}\frac{x}{y} = -0.1x$ Therefore, at $x=10$, the elasticity is $-1$.

0

For example, Take the car custom cleaned variable. Let's suppose it's represented by a dummy variable $C$ where $C=1$ implies the car is custom cleaned and $C=0$ means that it is not. Since C only takes on the values of 0 and 1, then $C^n=C$ for all natural numbers n. Thus, including both $C$ and $C^n$ in the regression leads to multicollinearity. However, ...

0

Rename $\beta \cdot \gamma = \delta$ and you have a regression in $\alpha$ and $\delta$ with $\epsilon_i$ denoting the model error.

1

I think we had some interaction on Stack Overflow a few days ago regarding F-stats. The concept of the F-test is fairly widespread in statistics, so it is not an innovation by Michael Newville (lmfit developer). Basically there are two types of F-tests to perform. Both of which deal with comparing two models, because an F is simply a ratio of chi-squares ...

-1

More initial thoughts rather than an answer... If $y=a+bx$ then $y_j-y_i=bx_j-bx_i$ $y_j-y_i=b\left(x_j-x_i\right)$

0

If you accept a fitting with a criterion different from the classical least mean square, the method is very simple : $$y-1=p_1(x-x^2y) +p_2(x^2-x^2y) +q_1(x^2y-xy)$$ From data $(x_1, y_1), (x_2, y_2), ... , (x_k, y_k), ... , (x_n, y_n)$, compute : $F_k=y_k-1$ $u_k=x_k-x_k^2y_k$ $v_k=x_k^2-x_k^2y_k$ $w_k=x_k^2y_k-x_ky_k$ Then, the linear regression : ...

0

Your problem is that the denominator can show roots. But this would not be the case if $$\Delta=-4 ({p_1}+{p_2})+{q_1}^2+4 {q_1}<0$$ I suppose that the easiest way to do the work would be to compute the sum of the squares of the residuals and try to minimize it using FMINCON with this nonlinear constraint (this seems to be part of the Optimization ...

1

It suffices to show that $X^{T}X (X^{T}X)^{g^{T}}$ is the identity matrix, which is true because you can interchange the generalized inverse and matrix transpose. Then the result follows by noticing that $X^{T}X$ is a symmetric matrix.

1

It is not true that $J(I-H)=0$ unless the rows of $J$, each of which is a row of $1$s, are in the row space of $X'$. That will certainly happen if one of the columns of $X$ is a column of $1$s, as happens in typical multiple regression problems. If you're not using the fact that one of the columns of $X$ is a column of $1$s, or failing that, that a column ...

0

I managed to solve it. $Var(e_{ij}) = Var(y_{ij}-\hat{y}_{ij}) = Var(y_{ij} - \bar{y}_{i.}) =$ $= Var(y_{ij}) + Var(\bar{y}_{ij}) - 2Cov(y_{ij},\bar{y}_{i.}) =$ $= Var(y_{ij}) + Var(\frac{1}{n_i}\sum_{i=1}^{n_i}{y_{ij}}) - 2Cov(y_{ij},\frac{1}{n_i}\sum_{i=1}^{n_i}{y_{ij}})$ $Var(y_{ij}) = \sigma^{2}$ $Var(\frac{1}{n_i}\sum_{i=1}^{n_i{y_{ij}}}) = ... 0 If there are no unit roots of the lag polynomials, then the integrated (I part) of an ARIMA process is equal to zero, thus we have an$ARIMA(1,0,1) = ARMA(1,1)$. As Alamos' first comment showed, there are no unit roots so it must be the case that we have an ARMA process. Hope this clears it up for you :) 1 The model you want to fit on the basis of$p$data points$(x_i,y_i)$is $$y=ax+b\sin(x)$$ To make life easier, define$z_i=\sin(x_i)$so the model becomes $$y=ax+bz$$ which corresponds to a multilinear regression with no intercept. What you want to minimize is $$F=\sum_{i=1}^p n_i(a x_i+b z_i-y_i)^2$$ So, the derivatives are $$\frac{dF}{da}=2\sum_{i=1}^p ... 1 The data (x_i,y_i) are given, so although it looks like you have a quadratic because of the x_i^2, in fact this is just a constant. You're solving for \alpha_0, \alpha_1 and \alpha_2, and the equation is linear in these terms. 1 You are looking for an equation y=mx+c. Ideally it would pass through all of the given points: that is, you would have -6m+c = -1, and similarly for other points. This is a set of four equations with two unknowns m,c. Its matrix representation is$$ \begin{pmatrix} -6 & 1\\ -2 & 1 \\ 1 & 1 \\ 7 & 1 \end{pmatrix} \begin{pmatrix} m \\ c ... 1 Here are some matrix algebra equations you can try. I'm suggesting this approach because you mentioned that there may be future cases where you have more than three points to work with and in such cases the linear regression solution is what you want. Starting with your equation for$s$, define the following quantities: $$\underline s = \begin{bmatrix} ... 0 You should probably use fmincon or lsqnonlin and take your derivatives by hand. Levenberg-Marquardt is normally the method of choice for non-linear least-squares, you can do it with lsqnonlin. Mathematical precision is measured by your cost function \Phi(a, b, c, d) 0 In multivariate analysis, the sign of the correlation is not necessarily equal to the sign of the coefficient in the equation. This is discussed here: http://stats.stackexchange.com/questions/34151/positive-correlation-and-negative-regressor-coefficient-sign It may help to consider the two-variable case. In this special case the sign of the correlation ... 1 Do know about log-log graph and semi-log graph? In log-log graph, both x and y axes are labeled as x and y but log(x) vs log(y) are plotted. They label x and y on the axes but scales are changed from 0,1,2,3... to 1,10,100,1000,.... So in your case, if you label x_1 on the x-axis, then use normal scaling 0,1,2,3,... and write on side of the graph about ... 1 As Paul said, the model is totally nonlinear with respect to its parameters and then nonlinear regression will be required. However, admit for the time being that you know a_4. So, you can write the model$$y -a_4 = \exp(a_1 \log (x) + a_2 \log (z) + a_3)\log(y -a_4) = {a_1 \log (x) + a_2 \log (z) + a_3}$$So, for given value of a_4 you can compute ... 1$$ e^{(a1 log (x) + a2 log (z) + a3)} + a_4 =e^{(log (x^{a_1}) + log (z^{a_2}) + a_3)} + a_4 =e^{(log (x^{a_1})}e^{+ log (z^{a_2})}e^{a_3)} + a_4=x^{a_1}z^{a_2}e^{a_3}+a_4$$Without a_4, you could do a linear regression in the log-transformed variables, but with it there is no way. So it would have to be a non-linear regression. 0 Matlab/Octave: To solve Ax=b, simply try: x = A\b; If you need the Cholesky factorization for some reason, the command is "chol". For example, G=chol(A); y=(G')\b; x=G\y; 0 You need to perform Cross Validation on \lambda. You also could approximate them using forward or backward subset selection. Theory wise, \lambda is the reduced cost of the objective function which comes from the shadow costs obtained from the dual optimization problem. Hence, nonlinear sensitivity analysis will help. This should be part of the output ... 2 The way you have things defined, it looks like you can read off the column vectors you seek from that matrix with respect to the linear transformation. The matrix of L with respect to \mathcal{S} and \mathcal{T} is one that is precisely$$\begin{bmatrix}[L\mathbf{v_1}]_{\mathcal{T}} & [L\mathbf{v_2}]_{\mathcal{T}} & ... 1 That's probably an Iverson bracket. 1 I am having trouble making sense of this problem. In your notation, the usual regression model is $$Y_i = a + bx_i + \epsilon_i,$$ where$\epsilon_i$are distributed$Norm(0, \sigma_\epsilon^2)$, for$i = 1, \dots, n.$A 95% confidence interval for the slope is $$b \pm t^* s_\epsilon \sqrt{1/S_{xx}},$$ where$t^* = 3.182\$ cuts off area .025 from the upper ...

Top 50 recent answers are included