The random variable X has probability density function f(x)={ax+bx2 , 0<x<1 . If E(X)=0.6 , find (a)P(X<1/2) and (b)var(x)
The solution of this problem is based on finding numerical values of parameters a and b in the PDF f(x).
Recall that the given PDF is zero over the domain R except that f(x) is non-zero over (0,1). Therefore all integration should be performed from -∞ to +∞ but for simplicity, will be carried out over the interval (0,1).
The area under all PDFs is exactly 1, so
∫f(x)dx (-∞ to +∞)=1
Integration of f(x) from 0 to 1 gives the equation 3a+2b=6.....(1)
Given E(X)=0.6 means
∫x*f(x)dx=0.6.
Performing the integration results in the equation 4a+3b=12*0.6=7.2 .... (2)
Solution of (1) and (2) gives
a=18/5, b=-12/5.
From the values of a and b,
(a) P(X<½)=∫f(x)dx from -∞ to ½
(b) Denoting E(X)=μ
var=∫(x-μ)²f(x)dx over domain R.
To find the values of a and b, we can use the fact that the probability density function (PDF) integrates to 1 over its support. In other words, we can solve the equation:
∫[0,1] (ax + bx^2) dx = 1
Integrating the PDF over the range [0, 1] gives us:
(a/2)x^2 + (b/3)x^3 | from 0 to 1 = 1
Substituting the limits:
(a/2)(1)^2 + (b/3)(1)^3 - (a/2)(0)^2 - (b/3)(0)^3 = 1
(a/2) + (b/3) = 1
Now, we need to find the values of a and b that satisfy this equation.
To find the value of a, we can use the fact that E(X) is the expected value of X, which is given as 0.6. We have:
E(X) = ∫[0,1] x * (ax + bx^2) dx
E(X) = a * ∫[0,1] x^2 dx + b * ∫[0,1] x^3 dx
E(X) = a * (1/3) + b * (1/4)
Since we know that E(X) = 0.6, we can substitute this value into the equation:
0.6 = a * (1/3) + b * (1/4)
This equation gives us one relation between a and b.
(a/2) + (b/3) = 1 ------> (1)
a/3 + b/4 = 0.6 ----------> (2)
Solving these two equations simultaneously, we can find the values of a and b.
To solve the above equations, we can multiply both sides of equation (2) by 12 to eliminate fractions:
4a + 3b = 7.2 --------> (3)
Next, multiply equation (1) by 6 to eliminate fractions:
3a + 2b = 6 ----------> (4)
Multiplying equation (4) by 2 gives us:
6a + 4b = 12 ---------> (5)
Now we can subtract equation (3) from equation (5):
(6a + 4b) - (4a + 3b) = 12 - 7.2
2a + b = 4.8
Now we can solve this equation for a and substitute the value of a back into equation (1) to find b:
Subtracting b from both sides:
2a = 4.8 - b
a = (4.8 - b)/2
Substitute the above expression for a into equation (1):
((4.8 - b)/2)/2 + b/3 = 1
2(4.8 - b)/2 + b/3 = 1
(4.8 - b) + (2/3)b = 1
4.8 - b + (2/3)b = 1
Multiply through by 3 to clear the fraction:
14.4 - 3b + 2b = 3
14.4 - b = 3
b = 14.4 - 3
b = 11.4
Substitute the value of b into equation (2) to find a:
a/3 + 11.4/4 = 0.6
a/3 + 11.4/4 = 0.6
Multiply through by 3 and 4 to clear the fractions:
4a + 34.2 = 7.2
4a = 7.2 - 34.2
4a = -27
a = -27/4
Now that we have found the values of a and b, we can answer the questions:
(a) P(X < 1/2) is the integral of f(x) from 0 to 1/2:
P(X < 1/2) = ∫[0,1/2] (ax + bx^2) dx
P(X < 1/2) = ∫[0,1/2] ((-27/4)x + (11.4)x^2) dx
(b) The variance of X, Var(X), can be calculated using the formula:
Var(X) = E(X^2) - [E(X)]^2
We already know E(X) = 0.6. To find E(X^2), we integrate x^2 * f(x) over the support:
E(X^2) = ∫[0,1] x^2 * (ax + bx^2) dx
E(X^2) = a ∫[0,1] x^3 dx + b ∫[0,1] x^4 dx
Once we have E(X^2), we can calculate Var(X) using the formula above.
To find the answers to (a) and (b), we need to calculate the values of the constants 'a' and 'b' in the given probability density function.
To find 'a' and 'b', we can use the fact that the probability density function integrates to 1 over its entire range. Therefore, we can set up the following equation:
∫[0 to 1] (ax + bx^2) dx = 1
Let's solve this equation to find the values of 'a' and 'b'.
∫[0 to 1] (ax + bx^2) dx = a∫[0 to 1] x dx + b∫[0 to 1] x^2 dx
Integrating each term separately:
[1/2(ax^2) from 0 to 1] + [1/3(bx^3) from 0 to 1] = a/2 + b/3
Now, we set this equal to 1 and solve for 'a' and 'b':
a/2 + b/3 = 1
To simplify, let's multiply through by 6 to get rid of the denominators:
3a + 2b = 6
Now we have a system of equations:
3a + 2b = 6 (equation 1)
E(X) = ∫[0 to 1] x (ax + bx^2) dx = 0.6 (equation 2)
Let's solve equation 2 to find the value of 'a'.
∫[0 to 1] x (ax + bx^2) dx = 0.6
Integrating each term separately:
[1/3(ax^3) from 0 to 1] + [1/4(bx^4) from 0 to 1] = a/3 + b/4
Substituting this back into equation 2:
0.6 = a/3 + b/4
Multiplying both sides by 12 to clear the denominators:
7a + 9b = 7.2 (equation 3)
Now, we have a system of equations:
3a + 2b = 6 (equation 1)
7a + 9b = 7.2 (equation 3)
Solving this system of equations will give us the values of 'a' and 'b'.
(a) To find P(X < 1/2), we need to integrate the probability density function from 0 to 1/2:
P(X < 1/2) = ∫[0 to 1/2] (ax + bx^2) dx
Substituting the values of 'a' and 'b' we found from the system of equations, we can calculate this value.
(b) To find the variance, we can use the formula for calculating variance:
Var(X) = E(X^2) - (E(X))^2
We already have the value of E(X) and we can calculate E(X^2) by integrating the product of the probability density function and x^2 over the range 0 to 1.
Substituting the values of 'a' and 'b' from the system of equations, we can calculate Var(X).