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Thread: SST = SSE + SSR, proof

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    SST = SSE + SSR, proof




    Hi,

    I've been trying to figure out for a long time how to derive the starting equation of the proof. I have no idea how to come up with it. It's the third equation on the picture in the attachement.

    Thank you very much for your help!
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    Re: SST = SSE + SSR, proof

    I read into this a bit last night. I believe you can add and subtract \hat{Y}_i. So, the right side of line 3 becomes: \sum_{i = 1}^n (\hat{Y}_i-\bar{Y}+Y_i-\hat{Y}_i)^2. I think you have a \hat{Y} in place of \bar{Y}. Then expand from here?

    This is the first line of the proof. I referenced https://web.njit.edu/~wguo/Math644_2...%201_part4.pdf
    Maybe we can finish the proof after fixing this small mistake?
    Last edited by Buckeye; 08-28-2016 at 09:29 AM.
    "I have discovered a truly remarkable proof of this theorem which this margin is too narrow to contain." Pierre de Fermat

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    Re: SST = SSE + SSR, proof


    It is perhaps best to use the deviation form of the regression function i.e.,

    y_{i}=\hat{y_{i}}+e_{i}.

    Squaring both sides and summing over the sample yields,

    \sum y_{i}^{2}=\sum \hat{y_{i}}^{2}+\sum e_{i}^{2}+2\sum \hat{y_{i}}e_{i}

    =\sum \hat{y_{i}}^{2}+\sum e_{i}^{2}

    =\hat{\beta _{1}^{2}}\sum x_{i}^{2}+\sum e_{i}^{2}

    because

    \sum \hat{y_{i}}e_{i}=0.

    As such, the various sums of squares above are:

    \sum y_{i}^{2} is the total variation (TSS),

    \sum \hat{y_{i}}^{2}=\hat{\beta _{1}^{2}}\sum x_{i}^{2} is the explained sum of squares from the regression (ESS),

    \sum e_{i}^{2} is the residual sum of squares (RSS).

    Thus, it follows that TSS=ESS+RSS.

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