Probability Density Functions: Mean and Variance Calculation

To determine the correct statement, we need to calculate the mean and variance for both probability density functions, \(f(x)\) and \(g(x)\). A probability density function (PDF) must satisfy two conditions: it must be non-negative for all \(x\), and the integral over its entire domain must equal 1.

Probability Density Function \(f(x)\) Analysis

The function \(f(x)\) is defined as:

\(f\left( x \right) = \left\{ {\begin{array}{} {\frac{x}{a} + 1}&{: - a \le x < 0}\\ { - \frac{x}{a} + 1}&{0 \le x \le a}\\ 0&{otherwise} \end{array}} \right.\)

Mean of \(f(x)\)

The mean (\(\mu_f\)) of a continuous probability density function \(f(x)\) is given by the formula:

\(\mu_f = E[X] = \int_{-\infty}^{\infty} x f(x) dx\)

For \(f(x)\), we integrate over the non-zero intervals:

\(\mu_f = \int_{-a}^{0} x \left(\frac{x}{a} + 1\right) dx + \int_{0}^{a} x \left(-\frac{x}{a} + 1\right) dx\)

\(\mu_f = \int_{-a}^{0} \left(\frac{x^2}{a} + x\right) dx + \int_{0}^{a} \left(-\frac{x^2}{a} + x\right) dx\)

Now, let's evaluate each integral:

• For the first integral (\(I_1\)):

  \(I_1 = \left[\frac{x^3}{3a} + \frac{x^2}{2}\right]_{-a}^{0}\)

  \(I_1 = \left(0 - \left(\frac{(-a)^3}{3a} + \frac{(-a)^2}{2}\right)\right)\)

  \(I_1 = -\left(-\frac{a^3}{3a} + \frac{a^2}{2}\right) = -\left(-\frac{a^2}{3} + \frac{a^2}{2}\right)\)

  \(I_1 = -\left(\frac{-2a^2 + 3a^2}{6}\right) = -\frac{a^2}{6}\)

• For the second integral (\(I_2\)):

  \(I_2 = \left[-\frac{x^3}{3a} + \frac{x^2}{2}\right]_{0}^{a}\)

  \(I_2 = \left(-\frac{a^3}{3a} + \frac{a^2}{2}\right) - 0\)

  \(I_2 = -\frac{a^2}{3} + \frac{a^2}{2} = \frac{-2a^2 + 3a^2}{6} = \frac{a^2}{6}\)

Combining these results:

\(\mu_f = I_1 + I_2 = -\frac{a^2}{6} + \frac{a^2}{6} = 0\)

So, the mean of \(f(x)\) is 0.

Variance of \(f(x)\)

The variance (\(\sigma_f^2\)) of a continuous probability density function is given by the formula:

\(\sigma_f^2 = E[X^2] - (E[X])^2\)

Since \(\mu_f = E[X] = 0\), the variance simplifies to \(\sigma_f^2 = E[X^2]\).

\(E[X^2] = \int_{-\infty}^{\infty} x^2 f(x) dx\)

\(E[X^2] = \int_{-a}^{0} x^2 \left(\frac{x}{a} + 1\right) dx + \int_{0}^{a} x^2 \left(-\frac{x}{a} + 1\right) dx\)

\(E[X^2] = \int_{-a}^{0} \left(\frac{x^3}{a} + x^2\right) dx + \int_{0}^{a} \left(-\frac{x^3}{a} + x^2\right) dx\)

Now, let's evaluate each integral:

• For the first integral (\(J_1\)):

  \(J_1 = \left[\frac{x^4}{4a} + \frac{x^3}{3}\right]_{-a}^{0}\)

  \(J_1 = \left(0 - \left(\frac{(-a)^4}{4a} + \frac{(-a)^3}{3}\right)\right)\)

  \(J_1 = -\left(\frac{a^4}{4a} - \frac{a^3}{3}\right) = -\left(\frac{a^3}{4} - \frac{a^3}{3}\right)\)

  \(J_1 = -\left(\frac{3a^3 - 4a^3}{12}\right) = -\left(-\frac{a^3}{12}\right) = \frac{a^3}{12}\)

• For the second integral (\(J_2\)):

  \(J_2 = \left[-\frac{x^4}{4a} + \frac{x^3}{3}\right]_{0}^{a}\)

  \(J_2 = \left(-\frac{a^4}{4a} + \frac{a^3}{3}\right) - 0\)

  \(J_2 = -\frac{a^3}{4} + \frac{a^3}{3} = \frac{-3a^3 + 4a^3}{12} = \frac{a^3}{12}\)

Combining these results:

\(\sigma_f^2 = E[X^2] = J_1 + J_2 = \frac{a^3}{12} + \frac{a^3}{12} = \frac{2a^3}{12} = \frac{a^3}{6}\)

So, the variance of \(f(x)\) is \(\frac{a^3}{6}\).

Probability Density Function \(g(x)\) Analysis

The function \(g(x)\) is defined as:

\(g\left( x \right) = \left\{ {\begin{array}{} { - \frac{x}{a}}&{:-a \le x \le 0}\\ {\frac{x}{a}}&{:0 \le x \le a}\\ 0&{:otherewise} \end{array}} \right.\)

Mean of \(g(x)\)

The mean (\(\mu_g\)) of a continuous probability density function \(g(x)\) is given by:

\(\mu_g = E[X] = \int_{-\infty}^{\infty} x g(x) dx\)

For \(g(x)\), we integrate over the non-zero intervals:

\(\mu_g = \int_{-a}^{0} x \left(-\frac{x}{a}\right) dx + \int_{0}^{a} x \left(\frac{x}{a}\right) dx\)

\(\mu_g = \int_{-a}^{0} \left(-\frac{x^2}{a}\right) dx + \int_{0}^{a} \left(\frac{x^2}{a}\right) dx\)

Now, let's evaluate each integral:

• For the first integral (\(K_1\)):

  \(K_1 = \left[-\frac{x^3}{3a}\right]_{-a}^{0}\)

  \(K_1 = \left(0 - \left(-\frac{(-a)^3}{3a}\right)\right) = -\left(-\frac{-a^3}{3a}\right) = -\left(\frac{a^2}{3}\right) = -\frac{a^2}{3}\)

• For the second integral (\(K_2\)):

  \(K_2 = \left[\frac{x^3}{3a}\right]_{0}^{a}\)

  \(K_2 = \left(\frac{a^3}{3a}\right) - 0 = \frac{a^2}{3}\)

Combining these results:

\(\mu_g = K_1 + K_2 = -\frac{a^2}{3} + \frac{a^2}{3} = 0\)

So, the mean of \(g(x)\) is 0.

Variance of \(g(x)\)

Since \(\mu_g = E[X] = 0\), the variance simplifies to \(\sigma_g^2 = E[X^2]\).

\(E[X^2] = \int_{-\infty}^{\infty} x^2 g(x) dx\)

\(E[X^2] = \int_{-a}^{0} x^2 \left(-\frac{x}{a}\right) dx + \int_{0}^{a} x^2 \left(\frac{x}{a}\right) dx\)

\(E[X^2] = \int_{-a}^{0} \left(-\frac{x^3}{a}\right) dx + \int_{0}^{a} \left(\frac{x^3}{a}\right) dx\)

Now, let's evaluate each integral:

• For the first integral (\(L_1\)):

  \(L_1 = \left[-\frac{x^4}{4a}\right]_{-a}^{0}\)

  \(L_1 = \left(0 - \left(-\frac{(-a)^4}{4a}\right)\right) = -\left(-\frac{a^4}{4a}\right) = -\left(-\frac{a^3}{4}\right) = \frac{a^3}{4}\)

• For the second integral (\(L_2\)):

  \(L_2 = \left[\frac{x^4}{4a}\right]_{0}^{a}\)

  \(L_2 = \left(\frac{a^4}{4a}\right) - 0 = \frac{a^3}{4}\)

Combining these results:

\(\sigma_g^2 = E[X^2] = L_1 + L_2 = \frac{a^3}{4} + \frac{a^3}{4} = \frac{2a^3}{4} = \frac{a^3}{2}\)

So, the variance of \(g(x)\) is \(\frac{a^3}{2}\).

Comparison of Means and Variances

Let's summarize our findings for the mean and variance of both probability density functions:

• Mean of \(f(x)\) (\(\mu_f\)) = 0
• Mean of \(g(x)\) (\(\mu_g\)) = 0

From this, we see that the mean of \(f(x)\) and the mean of \(g(x)\) are the same.

• Variance of \(f(x)\) (\(\sigma_f^2\)) = \(\frac{a^3}{6}\)
• Variance of \(g(x)\) (\(\sigma_g^2\)) = \(\frac{a^3}{2}\)

Since \(\frac{a^3}{6} \ne \frac{a^3}{2}\) (assuming \(a \ne 0\), which is implicit for the functions to be well-defined PDFs), the variance of \(f(x)\) and the variance of \(g(x)\) are different.

Conclusion

Based on our calculations, the mean of \(f(x)\) and \(g(x)\) are the same, while the variance of \(f(x)\) and \(g(x)\) are different. This matches the statement in option 2.