Let X be an exponential random variable with parameter λ=2. Find the values of the following. Use 'e' for the base of the natural logarithm (e.g., enter e^(-3) for e−3).
a) E[(3X+1)2]=
b) P(1≤X≤2)=
a)17/2
b)0.117
a) Let's break it down step by step!
First, let's find E[(3X+1)²]:
E[(3X+1)²] = E[9X² + 6X + 1]
Since X is an exponential random variable with parameter λ=2, we know that the mean of X, denoted as E[X], is equal to 1/λ = 1/2.
Now, let's use this information to find E[9X²] and E[6X]:
E[9X²] = 9(E[X])² = 9(1/2)² = 9/4
E[6X] = 6(E[X]) = 6(1/2) = 3
Finally, we can substitute these values into our equation:
E[(3X+1)²] = E[9X² + 6X + 1] = E[9X²] + E[6X] + E[1] = 9/4 + 3 + 1 = 9/4 + 4 = 25/4
Therefore, E[(3X+1)²] = 25/4.
b) To find P(1≤X≤2), we need to use the cumulative distribution function (CDF) of the exponential distribution.
The CDF of an exponential random variable X with parameter λ is given by F(x) = 1 - e^(-λx).
Let's use this formula:
P(1≤X≤2) = F(2) - F(1) = (1 - e^(-2*2)) - (1 - e^(-2*1)) = (1 - e^(-4)) - (1 - e^(-2))
Simplifying further:
P(1≤X≤2) = e^(-4) - e^(-2)
And there you have it! P(1≤X≤2) = e^(-4) - e^(-2).
To find the values, we'll use the properties of exponential random variables.
a) E[(3X+1)^2]:
Step 1: Recall that the expectation of a random variable is given by the formula:
E(X) = ∫[x * f(x)]dx
where f(x) is the probability density function (PDF) of the random variable.
Step 2: First, we need to find the PDF of the random variable (3X + 1)^2.
Let's define Y = (3X + 1)^2. Using the substitution method, we can express X in terms of Y:
Y = (3X + 1)^2
√Y = 3X + 1
X = (1/3)√Y - (1/3)
Next, let's find the PDF of Y.
To do this, let's calculate the cumulative distribution function (CDF) of Y and differentiate it.
CDF of Y:
F_Y(y) = P(Y ≤ y)
= P((3X + 1)^2 ≤ y)
= P((1/3)√Y - (1/3) ≤ X ≤ (1/3)√Y)
Since X is an exponential random variable with parameter λ = 2, we can plug in its PDF:
F_Y(y) = ∫[(1/3)√y - 1/3] * 2e^(-2x) dx
= ∫[(2/3)√y * e^(-2x) - (2/3)e^(-2x)] dx
= [-√y * e^(-2x) - (2/3)e^(-2x)]|_0^∞
= -√y * e^(-2∞) - (2/3)e^(-2∞) - (-(2/3)e^0)
= 0 - 0 - (-(2/3))(1)
= 2/3
Taking the derivative of the CDF, we obtain the PDF of Y:
f_Y(y) = d/dy (F_Y(y))
= 0 (for y ≤ 0) (since it's derivative of a constant)
= 2/3 (for y > 0)
Step 3: Now we can calculate the expected value:
E[(3X + 1)^2] = ∫[y * (2/3)]dy
= (2/3) * (1/2) * y^2 + c
= (1/3)y^2 + c
Therefore, E[(3X + 1)^2] = (1/3)y^2 + c.
b) P(1 ≤ X ≤ 2):
Step 1: To find the probability of X lying between 1 and 2, we need to calculate the cumulative distribution function (CDF) of X and subtract the CDF at 1 from the CDF at 2.
CDF of X:
F_X(x) = P(X ≤ x)
= ∫[0, x] 2e^(-2t) dt
= [-e^(-2t)]|[0, x]
= -e^(-2x) + 1
Step 2: Now, we can find the probability:
P(1 ≤ X ≤ 2) = F_X(2) - F_X(1)
= (-e^(-2(2)) + 1) - (-e^(-2(1)) + 1)
= (-e^(-4) + 1) - (-e^(-2) + 1)
= -e^(-4) + e^(-2)
Therefore, P(1 ≤ X ≤ 2) = -e^(-4) + e^(-2).
To find the expected value E[(3X+1)^2] and the probability P(1≤X≤2) for an exponential random variable, we need to use the properties and formulas of exponential distributions. Here's how to solve each part:
a) E[(3X+1)^2]:
The expected value of a random variable is the mean or average value of all the possible outcomes. For a continuous random variable like X, the expected value can be found using the formula:
E[g(X)] = ∫[−∞,∞] g(x) * f(x) dx
Where g(X) is the function of the random variable, f(x) is the probability density function (PDF) of the random variable, and the integral is taken over the entire range of possible values.
For an exponential random variable with parameter λ, the PDF is given by:
f(x) = λ * e^(-λx)
Now let's substitute the given values:
g(X) = (3X+1)^2
f(x) = 2 * e^(-2x)
To find E[(3X+1)^2], we need to calculate the integral:
E[(3X+1)^2] = ∫[0,∞] (3x+1)^2 * 2 * e^(-2x) dx
After simplifying and solving the integral, we get the expected value.
b) P(1≤X≤2):
The probability of an event occurring within a certain interval can be found by integrating the PDF over that interval:
P(a ≤ X ≤ b) = ∫[a,b] f(x) dx
In this case, we need to find P(1≤X≤2). Using the same PDF as before:
f(x) = 2 * e^(-2x)
we can calculate the integral:
P(1≤X≤2) = ∫[1,2] 2 * e^(-2x) dx
After solving the integral, we get the probability value.
So, to find the values of a) E[(3X+1)^2] and b) P(1≤X≤2), you need to calculate the integrals by substituting the given values in the formulas and evaluating them.