Problem 1 - Arbeitsbuch Mathematik für Ingenieure

Question

Gegeben ist eine Zufallsvariable \(X\) mit der Dichte \(f\) folgender Form $$ f(t)=\left\\{\begin{array}{ccc} 1+t & \text { für } & -1 \leq t<0 \\ 1-t & \text { für } & 0 \leq t<1 \\ 0 & \text { sonst } & \end{array}\right. $$ a) Bestimmen und skizzieren Sie die Verteilungsfunktion \(F\) von \(X\). b) Berechnen Sie \(P(X \leq-0.5)\) und \(P\left(X^{2} \geq 0.25\right)\). c) Geben Sie den Erwartungswert von \(X\) an.

Step-by-Step Solution

Answer

CDF is: \( F(t) =\begin{cases}0\, & t< -1\t+\frac{t^2}{2}+\frac{1}{2}\, & -1\le t< 0\frac{1}{2}+t-\frac{t^2}{2}\, & 0\le t<1\1\, & t\ge1\end{cases}\) \( P(X \leq -0.5) = \frac{1}{4} \), \( P(X^2 \geq 0.25) = \frac{3}{8} \), and \( E(X) = 0 \).

Step 1: Understand the PDF

The given probability density function (PDF) of the random variable \(X\) is defined piecewise as \[ f(t) = \begin{cases} 1 + t & -1 \text{for}\ t < 0 \ 1 - t & \text{for}\ 0 \text{define}\ t < 1 \ 0 & \text{else} \right. \] This defines the behavior of \(X\) over different intervals of \( t \).

Step 2: Find the CDF for \(-1 \leq t < 0\)

For \( -1 \leq t < 0 \), integrate the PDF:\[ F(t) = \int_{-1}^{t} (1 + u) du \] This results in: \[ F(t) = \left[ u + \frac{u^2}{2} \right] \bigg|_{-1}^{t} \] Evaluating the integral: \[ F(t) = \left( t + \frac{t^2}{2} \right) - \left( -1 + \frac{1}{2} \right) \] Simplifying: \[ F(t) = t + \frac{t^2}{2} + \frac{1}{2} \]

Step 3: Find the CDF for \(0 \leq t < 1\)

For \( 0 \leq t < 1 \), integrate the PDF: \[ F(t) = F(0) + \int_{0}^{t} (1 - u) du \] From the previous step, \( F(0) = \frac{1}{2} \). So, \[ F(t) = \frac{1}{2} + \left[ u - \frac{u^2}{2} \right] \bigg|_{0}^{t} \] Evaluating the integral: \[ F(t) = \frac{1}{2} + \left( t - \frac{t^2}{2} \right) \]

Step 4: Sketch the CDF

Plot the function: For \( -1 \leq t < 0 \), \( F(t) = t + \frac{t^2}{2} + \frac{1}{2} \) For \( 0 \leq t < 1 \), \( F(t) = \frac{1}{2} + t - \frac{t^2}{2} \). Ensure continuity at \( t = 0 \), where both expressions yield \( F(0) = \frac{1}{2} \).The CDF is then flat at \( F(t) = 1 \) for \( t \geq 1 \) and \( F(t) = 0 \) for \( t < -1 \).

Step 5: Compute \( P(X \leq -0.5) \)

Use the CDF found: \[ P(X \leq -0.5) = F(-0.5) \] Evaluate: \[ F(-0.5) = -0.5 + \frac{-0.5^2}{2} + \frac{1}{2} = \frac{1}{4} \]

Step 6: Compute \( P(X^2 \geq 0.25) \)

This event means \( P(|X| \geq 0.5) \): \[ P(X \leq -0.5) + P(X \geq 0.5) \] From CDF: \[ P(X \geq 0.5) = 1 - F(0.5) \]Evaluate: \[ F(0.5) = 0.5 - \frac{0.5^2}{2} + \frac{1}{2} = 0.875 \]So: \[ P(X^2 \geq 0.25) = \frac{1}{4} + (1 - 0.875) = \frac{3}{8} \]

Step 7: Compute the Expectation \(E(X)\)

Use linearity of expectation split integrals for \(-1\leq X \leq 1\):\[E(X)=\int_{-1}^{0} t (1+t) dt + \int_{0}^{1} t (1-t) dt \]Evaluate: \[ \left[\frac{t^2}{2} + \frac{t^3}{3}\right]_{-1}^0 + \left[\frac{t^2}{2} - \frac{t^3}{3}\right]_{0}^1 \]this simplifies to 0.

Key Concepts

Cumulative Distribution Function

The Cumulative Distribution Function (CDF) is a fundamental concept in probability theory. It describes the probability that a random variable, X, will take a value less than or equal to a particular number, t. To find the CDF, we integrate the probability density function (PDF) over the range of interest. In our problem, the CDF can be split into two regions because the PDF is defined differently for different ranges. For -1 ≤ t < 0, the CDF is found by integrating the PDF from -1 to t, while for 0 ≤ t < 1, we integrate the PDF from 0 to t and add the CDF value at t=0. This approach ensures the CDF is continuous and reflects the accumulative probability. Keep in mind that for any value of t outside the specified range, the CDF is 0 for t < -1 and 1 for t > 1, ensuring the function covers all possible values of X.

Expectation

Expectation, also known as the expected value or mean, provides a measure of the central tendency of a random variable. Think of it as the long-term average value of the variable. To calculate the expectation of X, denoted as E(X), we take the integral of the product of the variable and its PDF over the entire range of the variable. In the given exercise, we use \(E(X) = \int_{-1}^{0} t f(t) dt + \int_{0}^{1} t f(t) dt\). This breaks down the problem into smaller integrals because our PDF is defined piecewise. By computing these integrals separately and summing them, we find the overall mean of X, which often provides intuition about where the majority of values lie in probability distributions.

Probability

Probability is the measure of the likelihood that a certain event will occur, ranging from 0 (impossible event) to 1 (certain event). In our example, we're interested in specific probabilities, such as P(X ≤ -0.5) and P(X^2 ≥ 0.25). To find these, we use the CDF. For P(X ≤ -0.5), we directly evaluate the CDF at that point. For P(X^2 ≥ 0.25), we effectively look for the probability that X is either less than -0.5 or greater than 0.5, integrating the CDF to account for both intervals. This process helps determine the overall probability of complex events using the CDF, a powerful tool in probability.