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Hint: By Cauchy-Schwarz, we have $E\Big(\big|(X+Y)(X-Y)\big|\Big)\leq \sqrt{E\Big(|X+Y|^2\Big)\cdot E\Big(|X-Y|^2\Big)}$.

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begin quote If you know that $\sum_{i=1}^n e_i=0$, show that $\sum_{i=1}^n\epsilon_i=0$ where $e_i=Y_i-\hat{Y_i}$ and $\epsilon_i=Y_i-E[Y_i]$. I know that $$Y_i=B_0+B_1X_i+\epsilon_i$$ and $$E[Y_i]=B_0+B_1X_i$$ end quote Some assumptions are omitted here. Usually one says something like $Y_i = B_0+ B_1 X_i + \varepsilon_i$ where $B_0$, $B_1$, ...

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It's essentially due to Runge's Phenomenon for interpolating polynomials. Even without uncertainty, the behavior of higher order polynomials at their endpoints is very sensitive to the parameter values. A good way to think about it is that, in order to be a good fit to the sample data (i.e., not veer completely out of the range of the data), you need to ...

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You have usual regression model $$Y_i = bX_i + \varepsilon_i$$ but you can only measure $\tilde{Y}_i = Y_i+\delta_i$, with some measurement error $\delta_i$. Now the model becomes $$\tilde{Y}_i = bX_i + \varepsilon_i+\delta_i$$ and if $\varepsilon_i$ and $\delta_i$ are independent, the only thing that changes is that the variance of the error term ...

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Your linear model is $Y=\beta_1 X + \beta_0 + \varepsilon$, where $Y$ is the time spent and $X=1$ if senior, $0$ otherwise. In a simple linear model like that, we know that the least square estimates of $\beta_1$ and $\beta_0$ are given by: https://upload.wikimedia.org/math/e/5/b/e5b794026921e4b402ae7fb58b2fd7c3.png where $\bar{x}$ and $\bar{y}$ are the ...

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A quick and dirty way which works. If $B,D,F$ are known, the problem reduces to a linear regression. So, make a three dimension grid and for each triplet compute the sum of squares until you find a minimum. For the best triplet, recompute $A,C,D$ and start the nonlinear regression. Because of symmetry, you must not compute all points. Suppose that you ...

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The fitting of your data1 (from github) with the method based on an integral equation is shown below. I will try to joint a paper "Triple exponential.docx" where one can find the method of fitting (In French, but the equations are lisible on other languages) Latter on, a new data set was proposed (Data7 from github), with small scatter. The results below ...

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