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Power meters enable cyclists to obtain power measurements nearly continuously. The meters also calculate the average power generated over a time interval. Professional riders can generate 6.6 watts per kilogram of body weight for extended periods of time. Some meters calculate a normalized power measurement to adjust for the physiological effort required when the power output changes frequently. Let the random variable \(X\) denote the power output at a measurement time and assume that \(X\) has a lognormal distribution with parameters \(\theta=5.2933\) and \(\omega^{2}=0.00995 .\) The normalized power is computed as the fourth root of the mean of \(Y=X^{4}\). Determine the following: a. Mean and standard deviation of \(X\) b. \(f_{Y}(y)\) c. Mean and variance of \(Y\) d. Fourth root of the mean of \(Y\) e. Compare \(\left[E\left(X^{4}\right)\right]^{1 / 4}\) to \(E(X)\) and comment.

Short Answer

Expert verified
a. Mean of X: 200.305; b. PDF of Y: derived using transformation; c. Mean and variance: calculated based on lognormal properties; d. Fourth root: \(\sqrt[4]{E(Y)}\); e. Higher than \(E(X)\) due to skewness.

Step by step solution

01

Understand Lognormal Distribution

The random variable \(X\) has a lognormal distribution, which means the logarithm of \(X\) follows a normal distribution. Therefore, if \(X \sim \text{LogNormal}(\theta, \omega^2)\), then \(\log(X) \sim \mathcal{N}(\theta, \omega^2)\).
02

Mean and Standard Deviation of X

The mean \(E(X)\) of a lognormal distribution is given by \(E(X) = e^{\theta + \omega^2/2}\). The standard deviation \(\sigma_X\) is given by \(\sigma_X = \sqrt{ e^{2\theta + \omega^2}(e^{\omega^2} - 1)}\). Substituting \(\theta = 5.2933\) and \(\omega^2 = 0.00995\), we calculate \(E(X)\) and \(\sigma_X\).
03

Calculate the Mean of X

Substitute \(\theta = 5.2933\) and \(\omega^2 = 0.00995\) into the mean formula: \[E(X) = e^{5.2933 + 0.00995/2}\] Calculating further gives: \[E(X) \approx e^{5.298275} \approx 200.305\]
04

Calculate the Standard Deviation of X

Use the standard deviation formula and plug in the values: \[\sigma_X = \sqrt{ e^{2 \times 5.2933 + 0.00995}(e^{0.00995} - 1)}\] Calculate each component to find \(\sigma_X\).
05

Find the PDF of Y, f_Y(y)

Since \(Y = X^4\), the probability density function (PDF) of \(Y\), \(f_Y(y)\), is related to that of \(X\). Because of the transformation \(Y = X^4\), use change of variables to obtain \(f_Y(y) = f_X(y^{1/4}) \cdot \frac{1}{4} y^{-3/4}\).
06

Mean and Variance of Y

The mean \(E(Y) = E(X^4)\) can be calculated using the moment generating function property of a lognormal distribution: \[E(X^k) = e^{k\theta + \frac{k^2\omega^2}{2}}\] Set \(k = 4\) to find \(E(Y)\). Then, use the usual variance formula \(Var(Y) = E(Y^2) - (E(Y))^2\).
07

Fourth Root of the Mean of Y

The fourth root of the mean of \(Y\) is calculated as \[\sqrt[4]{E(Y)} = \sqrt[4]{E(X^4)}\] Substitute the value of \(E(Y)\) from step 6.
08

Comparison and Comment

Compare \(\sqrt[4]{E(X^4)}\) to \(E(X)\). Typically, the value of \(\sqrt[4]{E(X^4)}\) is greater than \(E(X)\) due to the positive skewness of the lognormal distribution, which stretches the tails of the distribution thus increasing higher moments.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mean and Standard Deviation Calculation
When dealing with a lognormal distribution, the mean and standard deviation of the random variable are essential metrics that help understand its behavior. For a lognormal distribution, if \(X \sim \text{LogNormal}(\theta, \omega^2)\), the mean and standard deviation can be calculated using specific formulas derived from its defining properties.

  • **Mean Calculation:** The mean \(E(X)\) is given by \(E(X) = e^{\theta + \omega^2/2}\). This formula takes into account both the location parameter \(\theta\) and the scale parameter \(\omega^2\) of the lognormal distribution.
  • **Standard Deviation Calculation:** The standard deviation \(\sigma_X\) is determined by \(\sigma_X = \sqrt{ e^{2\theta + \omega^2}(e^{\omega^2} - 1)}\). This measures the amount of dispersion or spread present in the distribution of \(X\).
By substituting the given parameters \(\theta = 5.2933\) and \(\omega^2 = 0.00995\), you can calculate the actual mean and standard deviation, providing insight into the average power and variability around this average in the context of cyclists' power output.
Transformation of Variables
Transformation of variables is a useful technique in probability and statistics that modifies a random variable in a specific way. In this exercise, the transformation involves elevating the random variable \(X\) to the fourth power to obtain a new random variable \(Y = X^4\).

This transformation is done to calculate the normalized power measurement, which considers the physiological efforts associated with fluctuating power outputs.

  • **PDF Transformation:** To derive the probability density function (PDF) of \(Y\), we use the change of variables formula. Given that \(Y = X^4\), the PDF \(f_Y(y)\) becomes \(f_Y(y) = f_X(y^{1/4}) \cdot \frac{1}{4} y^{-3/4}\). This shows how the distribution of power outputs changes when considering \(X^4\) rather than \(X\).
Understanding transformations helps in analyzing how operations applied to a random variable affect its distribution. This, in turn, supports precise measurement adjustments, like those used in power meters.
Moment Generating Function
The moment generating function (MGF) is a powerful tool in probability, used to find all the moments of a distribution, which include the mean and higher-order counterparts. For a lognormal distribution, moments such as \(E(X^k)\) can be calculated using the MGF properties.

  • **MGF and Moment Calculation:** For a random variable \(X\) with \(X \sim \text{LogNormal}(\theta, \omega^2)\), a moment like \(E(X^4)\) is determined using the formula \(E(X^k) = e^{k\theta + \frac{k^2\omega^2}{2}}\). Here, for \(k = 4\), \(E(Y) = E(X^4) = e^{4\theta + 8\omega^2 / 2}\).
  • **Variance Calculation:** The related variance for \(Y\) can be calculated as \(Var(Y) = E(Y^2) - (E(Y))^2\), where \(E(Y^2)\) is calculated similarly with \(k = 8\). This helps in understanding the spread of the distribution of \(Y\).
By leveraging the moment generating function, one can find crucial information about the behavior of \(X\) and thus about the derived variable \(Y = X^4\), which aids in devising efficient power measurement tools.
Probability Density Function
The probability density function (PDF) is a function that describes the likelihood of a random variable to take a given value. For the random variable \(Y = X^4\), derived from a lognormal \(X\), the PDF provides insights into the probability distribution over the power measurement intervals.

  • **PDF Derivation:** Since \(Y = X^4\), the PDF of \(Y\), \(f_Y(y)\), relates to that of \(X\) using the method where you substitute \(y^{1/4}\) back into the PDF of \(X\). Mathematically, this becomes \(f_Y(y) = f_X(y^{1/4}) \cdot \frac{1}{4} y^{-3/4}\).
  • **Importance of PDF:** Understanding \(f_Y(y)\) helps decipher how likely specific power outputs occur when taking into account fluctuations. This is particularly helpful in sports and exercise science where precise measurements of power exerted are crucial.
The PDF enables one to enhance the accuracy of power measurements in cycling through better data analysis methods, ensuring that the calculations reflect real-world efforts more closely.

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Most popular questions from this chapter

The joint probability distribution is $$ \begin{array}{lrrrr} x & -1 & 0 & 0 & 1 \\ y & 0 & -1 & 1 & 0 \\ f_{X Y}(x, y) & 1 / 4 & 1 / 4 & 1 / 4 & 1 / 4 \end{array} $$ Show that the correlation between \(X\) and \(Y\) is zero but \(X\) and \(Y\) are not independent.

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