4 added 5 characters in body edited Dec 10 '17 at 18:42 Anshu 8399 bronze badges Suppose that a random variable Y is defined as $$Y=\beta_0 + \beta_1x + \epsilon$$, where $$var(\epsilon)=\sigma^2$$, the error has zero mean, and is independent across observations. Suppose that we fit $$\hat{Y}=\hat{\beta_0}+\hat{\beta_1}x+\hat{\beta_2}x^2+\hat{\beta_3}x^3$$. I would like to show that $$var(\hat{\beta_0}+\hat{\beta_1}x)>var(\tilde{\beta_0}+\tilde{\beta_1}x)$$$$var(\hat{\beta_0}+\hat{\beta_1}x) \geq var(\tilde{\beta_0}+\tilde{\beta_1}x)$$, where $$(\tilde{\beta_0},\tilde{\beta_1})$$ are the estimates obtained by fitting the model correctly. One of my homework assignments asked me to show this for a specific dataset, but I think it should be true generally. It's easy to show that $$var(\hat{\beta_1}) \geq var(\tilde{\beta_1})$$ and so on, as this question seems to ask. But I'm finding it difficult to extend this result to $$var(\hat{\beta_0}+\hat{\beta_1}x) \geq var(\tilde{\beta_0}+\tilde{\beta_1}x)$$. We can break the terms apart into $$var(\hat{\beta_0}+\hat{\beta_1}x)=var(\hat{\beta_0})+x^2var(\hat{\beta_1})+2cov(\hat{\beta_0},\hat{\beta_1}x)$$, but how can I simplify the covariance term? I attempted to use the law of total covariance. Covariance is bilinear, and the estimated $$\beta_0$$'s are equivalent, so $$cov(\hat{\beta_0},\hat{\beta_1}x) \geq cov(\tilde{\beta_0},\tilde{\beta_1}x)$$ is equivalent to $$cov(\hat{\beta_0},(\hat{\beta_1}-\tilde{\beta_1})) \geq 0$$. By the law of total covariance, \begin{align*} Cov(\hat{\beta_0},(\hat{\beta_1}-\tilde{\beta_1})x) &= E(cov(\hat{\beta_0},(\hat{\beta_1}-\tilde{\beta_1})x|Y)) + cov(E(\hat{\beta_0}|Y),E(\hat{\beta_1}-\tilde{\beta_1}x|Y)) \\ &= E(cov(\bar{Y},(\hat{\beta_1}-\tilde{\beta_1})x|Y)) + cov(E(\bar{Y}|Y),E((\hat{\beta_1}-\tilde{\beta_1})x|Y)) \\ \end{align*} $$\bar{Y}|Y$$ is a constant, so its covariance with any other variable is zero. The first term equals zero. The second term, though, doesn't seem like it has to be positive or negative in general. Does anyone know if there's a way to complete the proof? My approaches have focused on reducing the covariance term to a separable function of the individual beta variances. Maybe there's another method? Suppose that a random variable Y is defined as $$Y=\beta_0 + \beta_1x + \epsilon$$, where $$var(\epsilon)=\sigma^2$$, the error has zero mean, and is independent across observations. Suppose that we fit $$\hat{Y}=\hat{\beta_0}+\hat{\beta_1}x+\hat{\beta_2}x^2+\hat{\beta_3}x^3$$. I would like to show that $$var(\hat{\beta_0}+\hat{\beta_1}x)>var(\tilde{\beta_0}+\tilde{\beta_1}x)$$, where $$(\tilde{\beta_0},\tilde{\beta_1})$$ are the estimates obtained by fitting the model correctly. One of my homework assignments asked me to show this for a specific dataset, but I think it should be true generally. It's easy to show that $$var(\hat{\beta_1}) \geq var(\tilde{\beta_1})$$ and so on, as this question seems to ask. But I'm finding it difficult to extend this result to $$var(\hat{\beta_0}+\hat{\beta_1}x) \geq var(\tilde{\beta_0}+\tilde{\beta_1}x)$$. We can break the terms apart into $$var(\hat{\beta_0}+\hat{\beta_1}x)=var(\hat{\beta_0})+x^2var(\hat{\beta_1})+2cov(\hat{\beta_0},\hat{\beta_1}x)$$, but how can I simplify the covariance term? I attempted to use the law of total covariance. Covariance is bilinear, and the estimated $$\beta_0$$'s are equivalent, so $$cov(\hat{\beta_0},\hat{\beta_1}x) \geq cov(\tilde{\beta_0},\tilde{\beta_1}x)$$ is equivalent to $$cov(\hat{\beta_0},(\hat{\beta_1}-\tilde{\beta_1})) \geq 0$$. By the law of total covariance, \begin{align*} Cov(\hat{\beta_0},(\hat{\beta_1}-\tilde{\beta_1})x) &= E(cov(\hat{\beta_0},(\hat{\beta_1}-\tilde{\beta_1})x|Y)) + cov(E(\hat{\beta_0}|Y),E(\hat{\beta_1}-\tilde{\beta_1}x|Y)) \\ &= E(cov(\bar{Y},(\hat{\beta_1}-\tilde{\beta_1})x|Y)) + cov(E(\bar{Y}|Y),E((\hat{\beta_1}-\tilde{\beta_1})x|Y)) \\ \end{align*} $$\bar{Y}|Y$$ is a constant, so its covariance with any other variable is zero. The first term equals zero. The second term, though, doesn't seem like it has to be positive or negative in general. Does anyone know if there's a way to complete the proof? My approaches have focused on reducing the covariance term to a separable function of the individual beta variances. Maybe there's another method? Suppose that a random variable Y is defined as $$Y=\beta_0 + \beta_1x + \epsilon$$, where $$var(\epsilon)=\sigma^2$$, the error has zero mean, and is independent across observations. Suppose that we fit $$\hat{Y}=\hat{\beta_0}+\hat{\beta_1}x+\hat{\beta_2}x^2+\hat{\beta_3}x^3$$. I would like to show that $$var(\hat{\beta_0}+\hat{\beta_1}x) \geq var(\tilde{\beta_0}+\tilde{\beta_1}x)$$, where $$(\tilde{\beta_0},\tilde{\beta_1})$$ are the estimates obtained by fitting the model correctly. One of my homework assignments asked me to show this for a specific dataset, but I think it should be true generally. It's easy to show that $$var(\hat{\beta_1}) \geq var(\tilde{\beta_1})$$ and so on, as this question seems to ask. But I'm finding it difficult to extend this result to $$var(\hat{\beta_0}+\hat{\beta_1}x) \geq var(\tilde{\beta_0}+\tilde{\beta_1}x)$$. We can break the terms apart into $$var(\hat{\beta_0}+\hat{\beta_1}x)=var(\hat{\beta_0})+x^2var(\hat{\beta_1})+2cov(\hat{\beta_0},\hat{\beta_1}x)$$, but how can I simplify the covariance term? I attempted to use the law of total covariance. Covariance is bilinear, and the estimated $$\beta_0$$'s are equivalent, so $$cov(\hat{\beta_0},\hat{\beta_1}x) \geq cov(\tilde{\beta_0},\tilde{\beta_1}x)$$ is equivalent to $$cov(\hat{\beta_0},(\hat{\beta_1}-\tilde{\beta_1})) \geq 0$$. By the law of total covariance, \begin{align*} Cov(\hat{\beta_0},(\hat{\beta_1}-\tilde{\beta_1})x) &= E(cov(\hat{\beta_0},(\hat{\beta_1}-\tilde{\beta_1})x|Y)) + cov(E(\hat{\beta_0}|Y),E(\hat{\beta_1}-\tilde{\beta_1}x|Y)) \\ &= E(cov(\bar{Y},(\hat{\beta_1}-\tilde{\beta_1})x|Y)) + cov(E(\bar{Y}|Y),E((\hat{\beta_1}-\tilde{\beta_1})x|Y)) \\ \end{align*} $$\bar{Y}|Y$$ is a constant, so its covariance with any other variable is zero. The first term equals zero. The second term, though, doesn't seem like it has to be positive or negative in general. Does anyone know if there's a way to complete the proof? My approaches have focused on reducing the covariance term to a separable function of the individual beta variances. Maybe there's another method? 3 converted strict inequalities into weak ones edited Dec 10 '17 at 18:15 Anshu 8399 bronze badges Suppose that a random variable Y is defined as $$Y=\beta_0 + \beta_1x + \epsilon$$, where $$var(\epsilon)=\sigma^2$$, the error has zero mean, and is independent across observations. Suppose that we fit $$\hat{Y}=\hat{\beta_0}+\hat{\beta_1}x+\hat{\beta_2}x^2+\hat{\beta_3}x^3$$. I would like to show that $$var(\hat{\beta_0}+\hat{\beta_1}x)>var(\tilde{\beta_0}+\tilde{\beta_1}x)$$, where $$(\tilde{\beta_0},\tilde{\beta_1})$$ are the estimates obtained by fitting the model correctly. One of my homework assignments asked me to show this for a specific dataset, but I think it should be true generally. It's easy to show that $$var(\hat{\beta_1})>var(\tilde{\beta_1})$$$$var(\hat{\beta_1}) \geq var(\tilde{\beta_1})$$ and so on, as this question seems to ask. But I'm finding it difficult to extend this result to $$var(\hat{\beta_0}+\hat{\beta_1}x)>var(\tilde{\beta_0}+\tilde{\beta_1}x)$$$$var(\hat{\beta_0}+\hat{\beta_1}x) \geq var(\tilde{\beta_0}+\tilde{\beta_1}x)$$. We can break the terms apart into $$var(\hat{\beta_0}+\hat{\beta_1}x)=var(\hat{\beta_0})+x^2var(\hat{\beta_1})+2cov(\hat{\beta_0},\hat{\beta_1}x)$$, but how can I simplify the covariance term? I attempted to use the law of total covariance. Covariance is bilinear, and the estimated $$\beta_0$$'s are equivalent, so $$cov(\hat{\beta_0},\hat{\beta_1}x) > cov(\tilde{\beta_0},\tilde{\beta_1}x)$$$$cov(\hat{\beta_0},\hat{\beta_1}x) \geq cov(\tilde{\beta_0},\tilde{\beta_1}x)$$ is equivalent to $$cov(\hat{\beta_0},(\hat{\beta_1}-\tilde{\beta_1})) \geq 0$$. By the law of total covariance, \begin{align*} Cov(\hat{\beta_0},(\hat{\beta_1}-\tilde{\beta_1})x) &= E(cov(\hat{\beta_0},(\hat{\beta_1}-\tilde{\beta_1})x|Y)) + cov(E(\hat{\beta_0}|Y),E(\hat{\beta_1}-\tilde{\beta_1}x|Y)) \\ &= E(cov(\bar{Y},(\hat{\beta_1}-\tilde{\beta_1})x|Y)) + cov(E(\bar{Y}|Y),E((\hat{\beta_1}-\tilde{\beta_1})x|Y)) \\ \end{align*} $$\bar{Y}|Y$$ is a constant, so its covariance with any other variable is zero. The first term equals zero. The second term, though, doesn't seem like it has to be positive or negative in general. Does anyone know if there's a way to complete the proof? My approaches have focused on reducing the covariance term to a separable function of the individual beta variances. Maybe there's another method? Suppose that a random variable Y is defined as $$Y=\beta_0 + \beta_1x + \epsilon$$, where $$var(\epsilon)=\sigma^2$$, the error has zero mean, and is independent across observations. Suppose that we fit $$\hat{Y}=\hat{\beta_0}+\hat{\beta_1}x+\hat{\beta_2}x^2+\hat{\beta_3}x^3$$. I would like to show that $$var(\hat{\beta_0}+\hat{\beta_1}x)>var(\tilde{\beta_0}+\tilde{\beta_1}x)$$, where $$(\tilde{\beta_0},\tilde{\beta_1})$$ are the estimates obtained by fitting the model correctly. One of my homework assignments asked me to show this for a specific dataset, but I think it should be true generally. It's easy to show that $$var(\hat{\beta_1})>var(\tilde{\beta_1})$$ and so on, as this question seems to ask. But I'm finding it difficult to extend this result to $$var(\hat{\beta_0}+\hat{\beta_1}x)>var(\tilde{\beta_0}+\tilde{\beta_1}x)$$. We can break the terms apart into $$var(\hat{\beta_0}+\hat{\beta_1}x)=var(\hat{\beta_0})+x^2var(\hat{\beta_1})+2cov(\hat{\beta_0},\hat{\beta_1}x)$$, but how can I simplify the covariance term? I attempted to use the law of total covariance. Covariance is bilinear, and the estimated $$\beta_0$$'s are equivalent, so $$cov(\hat{\beta_0},\hat{\beta_1}x) > cov(\tilde{\beta_0},\tilde{\beta_1}x)$$ is equivalent to $$cov(\hat{\beta_0},(\hat{\beta_1}-\tilde{\beta_1})) \geq 0$$. By the law of total covariance, \begin{align*} Cov(\hat{\beta_0},(\hat{\beta_1}-\tilde{\beta_1})x) &= E(cov(\hat{\beta_0},(\hat{\beta_1}-\tilde{\beta_1})x|Y)) + cov(E(\hat{\beta_0}|Y),E(\hat{\beta_1}-\tilde{\beta_1}x|Y)) \\ &= E(cov(\bar{Y},(\hat{\beta_1}-\tilde{\beta_1})x|Y)) + cov(E(\bar{Y}|Y),E((\hat{\beta_1}-\tilde{\beta_1})x|Y)) \\ \end{align*} $$\bar{Y}|Y$$ is a constant, so its covariance with any other variable is zero. The first term equals zero. The second term, though, doesn't seem like it has to be positive or negative in general. Does anyone know if there's a way to complete the proof? My approaches have focused on reducing the covariance term to a separable function of the individual beta variances. Maybe there's another method? Suppose that a random variable Y is defined as $$Y=\beta_0 + \beta_1x + \epsilon$$, where $$var(\epsilon)=\sigma^2$$, the error has zero mean, and is independent across observations. Suppose that we fit $$\hat{Y}=\hat{\beta_0}+\hat{\beta_1}x+\hat{\beta_2}x^2+\hat{\beta_3}x^3$$. I would like to show that $$var(\hat{\beta_0}+\hat{\beta_1}x)>var(\tilde{\beta_0}+\tilde{\beta_1}x)$$, where $$(\tilde{\beta_0},\tilde{\beta_1})$$ are the estimates obtained by fitting the model correctly. One of my homework assignments asked me to show this for a specific dataset, but I think it should be true generally. It's easy to show that $$var(\hat{\beta_1}) \geq var(\tilde{\beta_1})$$ and so on, as this question seems to ask. But I'm finding it difficult to extend this result to $$var(\hat{\beta_0}+\hat{\beta_1}x) \geq var(\tilde{\beta_0}+\tilde{\beta_1}x)$$. We can break the terms apart into $$var(\hat{\beta_0}+\hat{\beta_1}x)=var(\hat{\beta_0})+x^2var(\hat{\beta_1})+2cov(\hat{\beta_0},\hat{\beta_1}x)$$, but how can I simplify the covariance term? I attempted to use the law of total covariance. Covariance is bilinear, and the estimated $$\beta_0$$'s are equivalent, so $$cov(\hat{\beta_0},\hat{\beta_1}x) \geq cov(\tilde{\beta_0},\tilde{\beta_1}x)$$ is equivalent to $$cov(\hat{\beta_0},(\hat{\beta_1}-\tilde{\beta_1})) \geq 0$$. By the law of total covariance, \begin{align*} Cov(\hat{\beta_0},(\hat{\beta_1}-\tilde{\beta_1})x) &= E(cov(\hat{\beta_0},(\hat{\beta_1}-\tilde{\beta_1})x|Y)) + cov(E(\hat{\beta_0}|Y),E(\hat{\beta_1}-\tilde{\beta_1}x|Y)) \\ &= E(cov(\bar{Y},(\hat{\beta_1}-\tilde{\beta_1})x|Y)) + cov(E(\bar{Y}|Y),E((\hat{\beta_1}-\tilde{\beta_1})x|Y)) \\ \end{align*} $$\bar{Y}|Y$$ is a constant, so its covariance with any other variable is zero. The first term equals zero. The second term, though, doesn't seem like it has to be positive or negative in general. Does anyone know if there's a way to complete the proof? My approaches have focused on reducing the covariance term to a separable function of the individual beta variances. Maybe there's another method? 2 added 147 characters in body edited Dec 10 '17 at 2:20 Anshu 8399 bronze badges Suppose that a random variable Y is defined as $$Y=\beta_0 + \beta_1x + \epsilon$$, where $$var(\epsilon)=\sigma^2$$, the error has zero mean, and is independent across observations. Suppose that we fit $$\hat{Y}=\hat{\beta_0}+\hat{\beta_1}x+\hat{\beta_2}x^2+\hat{\beta_3}x^3$$. I would like to show that $$var(\hat{\beta_0}+\hat{\beta_1}x)>var(\tilde{\beta_0}+\tilde{\beta_1}x)$$, where $$(\tilde{\beta_0},\tilde{\beta_1})$$ are the estimates obtained by fitting the model correctly. One of my homework assignments asked me to show this for a specific dataset, but I think it should be true generally. It's easy to show that $$var(\hat{\beta_1})>var(\tilde{\beta_1})$$ and so on, as this question seems to ask. But I'm finding it difficult to extend this result to $$var(\hat{\beta_0}+\hat{\beta_1}x)>var(\tilde{\beta_0}+\tilde{\beta_1}x)$$. We can break the terms apart into $$var(\hat{\beta_0}+\hat{\beta_1}x)=var(\hat{\beta_0})+x^2var(\hat{\beta_1})+2cov(\hat{\beta_0},\hat{\beta_1}x)$$, but how can I simplify the covariance term? I attempted to use the law of total covariance. Covariance is bilinear, and the estimated $$\beta_0$$'s are equivalent, so $$cov(\hat{\beta_0},\hat{\beta_1}x) > cov(\tilde{\beta_0},\tilde{\beta_1}x)$$ is equivalent to $$cov(\hat{\beta_0},(\hat{\beta_1}-\tilde{\beta_1})) \geq 0$$. By the law of total covariance, \begin{align*} Cov(\hat{\beta_0},(\hat{\beta_1}-\tilde{\beta_1})x) &= E(cov(\hat{\beta_0},(\hat{\beta_1}-\tilde{\beta_1})x|Y)) + cov(E(\hat{\beta_0}|Y),E(\hat{\beta_1}-\tilde{\beta_1}x|Y)) \\ &= E(cov(\bar{Y},(\hat{\beta_1}-\tilde{\beta_1})x|Y)) + cov(E(\bar{Y}|Y),E((\hat{\beta_1}-\tilde{\beta_1})x|Y)) \\ \end{align*} $$\bar{Y}|Y$$ is a constant, so its covariance with any other variable is zero. The first term equals zero. The second term, though, doesn't seem like it has to be positive or negative in general. Does anyone know if there's a way to complete the proof? My approaches have focused on reducing the covariance term to a separable function of the individual beta variances. Maybe there's another method? Suppose that a random variable Y is defined as $$Y=\beta_0 + \beta_1x + \epsilon$$, where $$var(\epsilon)=\sigma^2$$, the error has zero mean, and is independent across observations. Suppose that we fit $$\hat{Y}=\hat{\beta_0}+\hat{\beta_1}x+\hat{\beta_2}x^2+\hat{\beta_3}x^3$$. I would like to show that $$var(\hat{\beta_0}+\hat{\beta_1}x)>var(\tilde{\beta_0}+\tilde{\beta_1}x)$$, where $$(\tilde{\beta_0},\tilde{\beta_1})$$ are the estimates obtained by fitting the model correctly. One of my homework assignments asked me to show this for a specific dataset, but I think it should be true generally. It's easy to show that $$var(\hat{\beta_1})>var(\tilde{\beta_1})$$ and so on, as this question seems to ask. But I'm finding it difficult to extend this result to $$var(\hat{\beta_0}+\hat{\beta_1}x)>var(\tilde{\beta_0}+\tilde{\beta_1}x)$$. We can break the terms apart into $$var(\hat{\beta_0}+\hat{\beta_1}x)=var(\hat{\beta_0})+x^2var(\hat{\beta_1})+2cov(\hat{\beta_0},\hat{\beta_1}x)$$, but how can I simplify the covariance term? I attempted to use the law of total covariance. Covariance is bilinear, and the estimated $$\beta_0$$'s are equivalent, so $$cov(\hat{\beta_0},\hat{\beta_1}x) > cov(\tilde{\beta_0},\tilde{\beta_1}x)$$ is equivalent to $$cov(\hat{\beta_0},(\hat{\beta_1}-\tilde{\beta_1})) \geq 0$$. By the law of total covariance, \begin{align*} Cov(\hat{\beta_0},(\hat{\beta_1}-\tilde{\beta_1})x) &= E(cov(\hat{\beta_0},(\hat{\beta_1}-\tilde{\beta_1})x|Y)) + cov(E(\hat{\beta_0}|Y),E(\hat{\beta_1}-\tilde{\beta_1}x|Y)) \\ &= E(cov(\bar{Y},(\hat{\beta_1}-\tilde{\beta_1})x|Y)) + cov(E(\bar{Y}|Y),E((\hat{\beta_1}-\tilde{\beta_1})x|Y)) \\ \end{align*} $$\bar{Y}|Y$$ is a constant, so its covariance with any other variable is zero. The first term equals zero. The second term, though, doesn't seem like it has to be positive or negative in general. Does anyone know if there's a way to complete the proof? Suppose that a random variable Y is defined as $$Y=\beta_0 + \beta_1x + \epsilon$$, where $$var(\epsilon)=\sigma^2$$, the error has zero mean, and is independent across observations. Suppose that we fit $$\hat{Y}=\hat{\beta_0}+\hat{\beta_1}x+\hat{\beta_2}x^2+\hat{\beta_3}x^3$$. I would like to show that $$var(\hat{\beta_0}+\hat{\beta_1}x)>var(\tilde{\beta_0}+\tilde{\beta_1}x)$$, where $$(\tilde{\beta_0},\tilde{\beta_1})$$ are the estimates obtained by fitting the model correctly. One of my homework assignments asked me to show this for a specific dataset, but I think it should be true generally. It's easy to show that $$var(\hat{\beta_1})>var(\tilde{\beta_1})$$ and so on, as this question seems to ask. But I'm finding it difficult to extend this result to $$var(\hat{\beta_0}+\hat{\beta_1}x)>var(\tilde{\beta_0}+\tilde{\beta_1}x)$$. We can break the terms apart into $$var(\hat{\beta_0}+\hat{\beta_1}x)=var(\hat{\beta_0})+x^2var(\hat{\beta_1})+2cov(\hat{\beta_0},\hat{\beta_1}x)$$, but how can I simplify the covariance term? I attempted to use the law of total covariance. Covariance is bilinear, and the estimated $$\beta_0$$'s are equivalent, so $$cov(\hat{\beta_0},\hat{\beta_1}x) > cov(\tilde{\beta_0},\tilde{\beta_1}x)$$ is equivalent to $$cov(\hat{\beta_0},(\hat{\beta_1}-\tilde{\beta_1})) \geq 0$$. By the law of total covariance, \begin{align*} Cov(\hat{\beta_0},(\hat{\beta_1}-\tilde{\beta_1})x) &= E(cov(\hat{\beta_0},(\hat{\beta_1}-\tilde{\beta_1})x|Y)) + cov(E(\hat{\beta_0}|Y),E(\hat{\beta_1}-\tilde{\beta_1}x|Y)) \\ &= E(cov(\bar{Y},(\hat{\beta_1}-\tilde{\beta_1})x|Y)) + cov(E(\bar{Y}|Y),E((\hat{\beta_1}-\tilde{\beta_1})x|Y)) \\ \end{align*} $$\bar{Y}|Y$$ is a constant, so its covariance with any other variable is zero. The first term equals zero. The second term, though, doesn't seem like it has to be positive or negative in general. Does anyone know if there's a way to complete the proof? My approaches have focused on reducing the covariance term to a separable function of the individual beta variances. Maybe there's another method? 1 asked Dec 9 '17 at 18:57 Anshu 8399 bronze badges