Problem 48
Question
Let \(X\) be exponentially distributed with parameter \(\lambda\). Find \(\operatorname{var}(X)\)
Step-by-Step Solution
Verified Answer
The variance of \(X\) is \(\frac{1}{\lambda^2}\).
1Step 1: Recall the Definition of Variance
The variance of a random variable is defined as the expected value of the squared deviation of the variable from its mean, expressed as \(\operatorname{var}(X) = E(X^2) - (E(X))^2\).
2Step 2: Determine the Mean of an Exponential Distribution
For an exponentially distributed random variable with parameter \(\lambda\), the mean, which is \(E(X)\), is given by \(\frac{1}{\lambda}\).
3Step 3: Compute the Second Moment
The second moment, \(E(X^2)\), of an exponential distribution with parameter \(\lambda\) is given by \(\frac{2}{\lambda^2}\). This is a standard result derived from the exponential distribution's density function.
4Step 4: Calculate the Variance
Substitute the values of \(E(X)\) and \(E(X^2)\) into the variance formula: \[ \operatorname{var}(X) = E(X^2) - (E(X))^2 = \frac{2}{\lambda^2} - \left(\frac{1}{\lambda}\right)^2 = \frac{2}{\lambda^2} - \frac{1}{\lambda^2} = \frac{1}{\lambda^2}.\]
Key Concepts
Expected ValueSecond MomentExponential Distribution Parameter
Expected Value
The expected value, often denoted as \(E(X)\), of a random variable is a fundamental concept in probability. It represents the average or mean value that the variable takes after many iterations or trials.
In the realm of exponential distributions, which are continuous probability distributions, the expected value takes on a special significance. When you have a random variable \(X\) that follows an exponential distribution with parameter \(\lambda\), its expected value is expressed as \(\frac{1}{\lambda}\). This means the average time until an event occurs, or the average number of successes over a long period, is directly related to the inverse of \(\lambda\).
Think of it this way: if \(\lambda\) is high, the exponential distribution is steep, indicating that the event is expected to happen quickly, resulting in a smaller expected value. Conversely, with a smaller \(\lambda\), the distribution is spread out over time, suggesting events occur less frequently, thus leading to a larger expected value.
In the realm of exponential distributions, which are continuous probability distributions, the expected value takes on a special significance. When you have a random variable \(X\) that follows an exponential distribution with parameter \(\lambda\), its expected value is expressed as \(\frac{1}{\lambda}\). This means the average time until an event occurs, or the average number of successes over a long period, is directly related to the inverse of \(\lambda\).
Think of it this way: if \(\lambda\) is high, the exponential distribution is steep, indicating that the event is expected to happen quickly, resulting in a smaller expected value. Conversely, with a smaller \(\lambda\), the distribution is spread out over time, suggesting events occur less frequently, thus leading to a larger expected value.
Second Moment
The second moment of a random variable is a concept that builds upon the idea of expectation. While the expected value gives us a first measure central tendency, the second moment provides insights into the variability around this average.
For an exponentially distributed random variable \(X\) with parameter \(\lambda\), the second moment \(E(X^2)\) is given by \(\frac{2}{\lambda^2}\). This metric helps us understand how values of \(X\) are typically distributed around the mean.
The calculation of the second moment for an exponential distribution is derived from its probability density function. The density function gives this distribution its characteristic "memoryless" property, meaning the expectation doesn't change over time. Understanding the second moment is crucial for determining the variance, which quantifies the spread of the distribution.
For an exponentially distributed random variable \(X\) with parameter \(\lambda\), the second moment \(E(X^2)\) is given by \(\frac{2}{\lambda^2}\). This metric helps us understand how values of \(X\) are typically distributed around the mean.
The calculation of the second moment for an exponential distribution is derived from its probability density function. The density function gives this distribution its characteristic "memoryless" property, meaning the expectation doesn't change over time. Understanding the second moment is crucial for determining the variance, which quantifies the spread of the distribution.
Exponential Distribution Parameter
The parameter \(\lambda\) in an exponential distribution is pivotal. It dictates the shape and rate of the distribution. With \(\lambda\) being a positive real number, it defines how fast or slow the exponential distribution decays.
In simpler terms, \(\lambda\) can be seen as the rate at which the events occur. A higher \(\lambda\) means more frequent events, with a sharper drop-off in probabilities for longer wait times. Conversely, a lower value suggests that events are less frequent, extending the probability of observing larger values.
Understanding \(\lambda\) aids in grasping other related characteristics of exponential distributions, such as the expected value and variance. For instance, knowing the expected value is \(\frac{1}{\lambda}\) and variance is \(\frac{1}{\lambda^2}\), you can see how \(\lambda\) directly affects the central tendency and spread of the distribution. It helps us model real-world phenomena where events occur continuously and independently over time, such as radioactive decay or time between arrivals at a service center.
In simpler terms, \(\lambda\) can be seen as the rate at which the events occur. A higher \(\lambda\) means more frequent events, with a sharper drop-off in probabilities for longer wait times. Conversely, a lower value suggests that events are less frequent, extending the probability of observing larger values.
Understanding \(\lambda\) aids in grasping other related characteristics of exponential distributions, such as the expected value and variance. For instance, knowing the expected value is \(\frac{1}{\lambda}\) and variance is \(\frac{1}{\lambda^2}\), you can see how \(\lambda\) directly affects the central tendency and spread of the distribution. It helps us model real-world phenomena where events occur continuously and independently over time, such as radioactive decay or time between arrivals at a service center.
Other exercises in this chapter
Problem 47
Thirteen cards are drawn at random without replacement from a standard deck of 52 cards. What is the probability that all are red?
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an urn contains \(K\) green and \(N-K\) blue balls. (a) You take \(n\) balls out of the urn. Find the probability that \(k\) of the \(n\) balls are green. (b) Y
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Four cards are drawn at random without replacement from a standard deck of 52 cards. What is the probability that all are of different suits?
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Suppose that the lifetime of a battery is exponentially distributed with an average life span of three months. What is the probability that the battery will las
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