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Prove the following theorem: Given a density forecast $f\left(y_{T+h} \mid \mat...
Aug 20, 2024
Solution by Steps
step 1
To minimize the expected absolute loss, we need to find the point forecast y^T+h\hat{y}_{T+h} that minimizes the expected loss function, which is given by the integral of the absolute loss: E[L(y^,y)]=∫−∞+∞∣y^−y∣f(yT+h∣IT,θ)dyE[L(\hat{y}, y)] = \int_{-\infty}^{+\infty} | \hat{y} - y | f(y_{T+h} | \mathcal{I}_{T}, \boldsymbol{\theta}) dy
step 2
The absolute loss function L(y^,y)=∣y^−y∣L(\hat{y}, y) = | \hat{y} - y | can be split into two cases based on whether yy is less than or greater than y^\hat{y}. Thus, we can express the expected loss as: E[L(y^,y)]=∫−∞y^(y^−y)f(yT+h∣IT,θ)dy+∫y^+∞(y−y^)f(yT+h∣IT,θ)dyE[L(\hat{y}, y)] = \int_{-\infty}^{\hat{y}} (\hat{y} - y) f(y_{T+h} | \mathcal{I}_{T}, \boldsymbol{\theta}) dy + \int_{\hat{y}}^{+\infty} (y - \hat{y}) f(y_{T+h} | \mathcal{I}_{T}, \boldsymbol{\theta}) dy
step 3
To find the optimal y^\hat{y}, we take the derivative of the expected loss with respect to y^\hat{y} and set it to zero. This leads to the condition that the probability mass on either side of y^\hat{y} must be equal, which is mathematically expressed as: ∫−∞y^f(yT+h∣IT,θ)dy=12\int_{-\infty}^{\hat{y}} f(y_{T+h} | \mathcal{I}_{T}, \boldsymbol{\theta}) dy = \frac{1}{2}
step 4
The unique value y^=mT+h\hat{y} = m_{T+h} that satisfies this condition is the median of the distribution, confirming that to minimize absolute loss, the point forecast should be set to the median: y^T+h=mT+h\hat{y}_{T+h} = m_{T+h}
Answer
The point forecast that minimizes expected absolute loss is the median, y^T+h=mT+h\hat{y}_{T+h} = m_{T+h}.
Key Concept
The median minimizes the expected absolute loss in a density forecast.
Explanation
The median is the value that divides the probability distribution into two equal halves, ensuring that the expected loss is minimized when the forecast is set to this value.
show proof with steps
To prove that the point forecast which minimizes expected absolute loss is the median mT+h m_{T+h} , we will follow these steps: Step 1: Define the Expected Absolute Loss The expected absolute loss is given by the integral of the absolute loss function over the density forecast: E[L(y^,y)]=∫−∞+∞∣y^−y∣f(yT+h∣IT,θ) dy E[L(\hat{y}, y)] = \int_{-\infty}^{+\infty} |\hat{y} - y| f(y_{T+h} \mid \mathcal{I}_{T}, \boldsymbol{\theta}) \, dy Step 2: Analyze the Absolute Loss Function The absolute loss function L(y^,y)=∣y^−y∣ L(\hat{y}, y) = |\hat{y} - y| can be split into two cases based on the value of y y relative to y^ \hat{y} : 1. If y < \hat{y} , then ∣y^−y∣=y^−y |\hat{y} - y| = \hat{y} - y . 2. If y≥y^ y \geq \hat{y} , then ∣y^−y∣=y−y^ |\hat{y} - y| = y - \hat{y} . Step 3: Split the Integral We can split the expected loss into two parts: E[L(y^,y)]=∫−∞y^(y^−y)f(yT+h∣IT,θ) dy+∫y^+∞(y−y^)f(yT+h∣IT,θ) dy E[L(\hat{y}, y)] = \int_{-\infty}^{\hat{y}} (\hat{y} - y) f(y_{T+h} \mid \mathcal{I}_{T}, \boldsymbol{\theta}) \, dy + \int_{\hat{y}}^{+\infty} (y - \hat{y}) f(y_{T+h} \mid \mathcal{I}_{T}, \boldsymbol{\theta}) \, dy Step 4: Simplify the Expected Loss Let’s denote the cumulative distribution function (CDF) of the density forecast as F(y)=∫−∞yf(yT+h∣IT,θ) dy F(y) = \int_{-\infty}^{y} f(y_{T+h} \mid \mathcal{I}_{T}, \boldsymbol{\theta}) \, dy . The expected loss can be rewritten as: E[L(y^,y)]=y^F(y^)−∫−∞y^yf(yT+h∣IT,θ) dy+∫y^+∞yf(yT+h∣IT,θ) dy−y^(1−F(y^)) E[L(\hat{y}, y)] = \hat{y} F(\hat{y}) - \int_{-\infty}^{\hat{y}} y f(y_{T+h} \mid \mathcal{I}_{T}, \boldsymbol{\theta}) \, dy + \int_{\hat{y}}^{+\infty} y f(y_{T+h} \mid \mathcal{I}_{T}, \boldsymbol{\theta}) \, dy - \hat{y} (1 - F(\hat{y})) Step 5: Differentiate the Expected Loss To find the value of y^ \hat{y} that minimizes the expected loss, we differentiate E[L(y^,y)] E[L(\hat{y}, y)] with respect to y^ \hat{y} and set the derivative to zero: ddy^E[L(y^,y)]=F(y^)−(1−F(y^))=0 \frac{d}{d\hat{y}} E[L(\hat{y}, y)] = F(\hat{y}) - (1 - F(\hat{y})) = 0 This implies: F(y^)=12 F(\hat{y}) = \frac{1}{2} Step 6: Conclusion The condition F(y^)=12 F(\hat{y}) = \frac{1}{2} indicates that y^ \hat{y} is the median mT+h m_{T+h} of the distribution. Therefore, the point forecast that minimizes expected absolute loss is: y^T+h=mT+h \hat{y}_{T+h} = m_{T+h} This completes the proof that the median minimizes the expected absolute loss when using the absolute loss function.
write the proof in latex, every step of the way and show how you simplify it
Solution by Steps
step 1
To minimize the expected absolute loss, we need to find the point forecast hatyT+h\\hat{y}_{T+h} that minimizes the expected loss function, which is given by the integral of the absolute loss: E[L(haty,y)]=int−infty+infty∣haty−y∣f(yT+h∣mathcalIT,boldsymboltheta)dyE[L(\\hat{y}, y)] = \\int_{-\\infty}^{+\\infty} | \\hat{y} - y | f(y_{T+h} | \\mathcal{I}_{T}, \\boldsymbol{\\theta}) dy
step 2
The absolute loss function can be split into two cases based on whether yy is less than or greater than haty\\hat{y}: E[L(haty,y)]=int−inftyhaty(haty−y)f(yT+h∣mathcalIT,boldsymboltheta)dy+inthaty+infty(y−haty)f(yT+h∣mathcalIT,boldsymboltheta)dyE[L(\\hat{y}, y)] = \\int_{-\\infty}^{\\hat{y}} (\\hat{y} - y) f(y_{T+h} | \\mathcal{I}_{T}, \\boldsymbol{\\theta}) dy + \\int_{\\hat{y}}^{+\\infty} (y - \\hat{y}) f(y_{T+h} | \\mathcal{I}_{T}, \\boldsymbol{\\theta}) dy
step 3
To find the optimal haty\\hat{y}, we differentiate the expected loss with respect to haty\\hat{y} and set the derivative to zero. This leads to the condition: int−inftyhatyf(yT+h∣mathcalIT,boldsymboltheta)dy=inthaty+inftyf(yT+h∣mathcalIT,boldsymboltheta)dy\\int_{-\\infty}^{\\hat{y}} f(y_{T+h} | \\mathcal{I}_{T}, \\boldsymbol{\\theta}) dy = \\int_{\\hat{y}}^{+\\infty} f(y_{T+h} | \\mathcal{I}_{T}, \\boldsymbol{\\theta}) dy
step 4
The above condition implies that the point haty\\hat{y} must satisfy: int−inftymT+hf(yT+h∣mathcalIT,boldsymboltheta)dy=frac12\\int_{-\\infty}^{m_{T+h}} f(y_{T+h} | \\mathcal{I}_{T}, \\boldsymbol{\\theta}) dy = \\frac{1}{2}, which is the definition of the median mT+hm_{T+h}. Thus, the point forecast that minimizes expected absolute loss is the median: hatyT+h=mT+h\\hat{y}_{T+h} = m_{T+h}
Answer
The point forecast that minimizes expected absolute loss is the median mT+hm_{T+h}.
Key Concept
The median minimizes the expected absolute loss when using the absolute loss function.
Explanation
The proof shows that the optimal forecast is the median, as it balances the probabilities of being above and below this value, thus minimizing the expected loss.
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