Break a stick of length 1 at a uniform location and then choose either of the two
parts with equal probabilities. If X denotes the length of the part you choose, compute E[X]
and var(X).
To compute the expected value (E[X]) and variance (Var(X)) for the length of the part chosen after breaking a stick of length 1 at a uniform location, we can use some basic principles of probability.
Step 1: Understanding the problem
We are breaking a stick of length 1 at a random point, resulting in two parts. We then choose one of these two parts with equal probabilities.
Step 2: Calculate the probability density function (PDF)
Let's assume that the break point is at a distance x from one end of the stick, where 0 ≤ x ≤ 1. The PDF of x can be represented as f(x) = 1 for 0 ≤ x ≤ 1.
Step 3: Calculate the length of the chosen part
If the break point is at a distance x from one end, the length of the part chosen would be X = max(x, 1 - x).
Step 4: Calculate E[X]
The expected value (E[X]) of a random variable X can be found by integrating the product of X and its PDF over the entire range of values X can take.
E[X] = ∫ [0,1] max(x, 1 - x) * f(x) dx
= ∫ [0,1] max(x, 1 - x) dx
To solve this integral, we break it down into two parts:
Part 1: x ≤ 1 -x (0 ≤ x ≤ 0.5)
In this case, max(x, 1 - x) = 1 - x.
∫ [0,0.5] (1 - x) dx = ([x - (x^2)/2] from 0 to 0.5)
= [(0.5 - (0.5^2)/2) - (0 - (0^2)/2)]
= [0.5 - 0.125 - 0 + 0]
= 0.375
Part 2: x > 1 -x (0.5 < x ≤ 1)
In this case, max(x, 1 - x) = x.
∫ [0.5,1] x dx = [(x^2)/2] from 0.5 to 1
= [(1^2)/2 - (0.5^2)/2]
= [0.5 - 0.125]
= 0.375
Therefore, E[X] = 0.375 + 0.375 = 0.75
Step 5: Calculate Var(X)
The variance (Var(X)) of a random variable X measures the spread or dispersion of the values of X around the expected value E[X]. It can be calculated using the formula:
Var(X) = E[X^2] - (E[X])^2
To calculate E[X^2], we substitute X = max(x, 1 - x) as before and square it:
E[X^2] = ∫ [0,1] (max(x, 1 - x))^2 * f(x) dx
= ∫ [0,1] (max(x, 1 - x))^2 dx
Solving the integral, we get:
Part 1: x ≤ 1 -x (0 ≤ x ≤ 0.5)
In this case, (max(x, 1 - x))^2 = (1 - x)^2.
∫ [0,0.5] (1 - x)^2 dx = (x - (2x^2)/3 + (x^3)/3) from 0 to 0.5
= (0.5 - (2 * 0.5^2)/3 + (0.5^3)/3) - (0 - 0 + 0)
= 0.125
Part 2: x > 1 -x (0.5 < x ≤ 1)
In this case, (max(x, 1 - x))^2 = x^2.
∫ [0.5,1] x^2 dx = [(x^3)/3] from 0.5 to 1
= [(1^3)/3 - (0.5^3)/3]
= 0.125
Therefore, E[X^2] = 0.125 + 0.125 = 0.25
Now, we can calculate Var(X):
Var(X) = E[X^2] - (E[X])^2
= 0.25 - (0.75)^2
= 0.25 - 0.5625
= 0.1875
Thus, the expected value E[X] is 0.75 and the variance Var(X) is 0.1875 for the length of the part chosen after breaking a stick of length 1 at a uniform location.
To compute the expected value and variance of the length of the part you choose, we can use the properties of probability and expectation.
First, let's consider the probability distribution of X.
When the stick is broken at a uniform location, any point on the stick is equally likely to be chosen. Therefore, the probability of choosing a part with a length less than or equal to x is equal to the length x of that part.
Let's define the random variable Y as the location where the stick is broken, where 0 ≤ Y ≤ 1. Since the location is chosen uniformly, the probability density function (PDF) of Y is a constant 1 over the interval [0, 1].
Now, consider X as the length of the part you choose. We can express X in terms of Y as follows:
X = min(Y, 1 - Y)
Next, let's find the cumulative distribution function (CDF) of X, denoted as F(x). The CDF gives us the probability that X is less than or equal to x.
F(x) = P(X ≤ x)
To compute F(x), we need to find the probability that either Y is less than or equal to x or 1 - Y is less than or equal to x.
Case 1: Y ≤ x
In this case, X = Y, and the PDF of X is simply the PDF of Y.
Since Y is uniformly distributed between 0 and 1, the probability of Y being less than or equal to x is x.
Therefore, the probability of X being less than or equal to x when Y is less than or equal to x is x.
Case 2: 1 - Y ≤ x
In this case, X = 1 - Y. To find the probability of X being less than or equal to x when 1 - Y is less than or equal to x, we can rearrange the inequality:
1 - Y ≤ x => Y ≥ 1 - x
Since Y is uniformly distributed between 0 and 1, the probability of Y being greater than or equal to 1 - x is (1 - x).
Therefore, the probability of X being less than or equal to x when 1 - Y is less than or equal to x is 1 - x.
To find the overall probability that X is less than or equal to x, we can add the probabilities from the two cases:
F(x) = P(X ≤ x) = x + (1 - x) = 1
Now, let's find the probability density function (PDF) of X, denoted as f(x), by taking the derivative of the CDF.
f(x) = d/dx F(x)
Since the CDF is a constant 1, the derivative is zero for any value of x.
Therefore, the PDF of X is:
f(x) = 0 for 0 ≤ x ≤ 1
Now, we can compute the expected value E[X] and variance var(X) using the PDF of X.
Expected Value (E[X]):
E[X] = ∫ x * f(x) dx for -∞ ≤ x ≤ ∞
Since the PDF of X is zero for any value of x, the integral evaluates to zero. Therefore, the expected value of X is zero:
E[X] = 0
Variance (var(X)):
var(X) = ∫ (x - E[X])^2 * f(x) dx for -∞ ≤ x ≤ ∞
Now, since E[X] is zero, the integral simplifies to:
var(X) = ∫ x^2 * f(x) dx for -∞ ≤ x ≤ ∞
Since the PDF of X is zero for any value of x, the integral evaluates to zero. Therefore, the variance of X is zero:
var(X) = 0
In conclusion, the expected value of X is zero (E[X] = 0) and the variance of X is also zero (var(X) = 0). This means that regardless of the position at which the stick is broken, the length of the part you choose will have no variability as it will always be zero.