The random variable X has a PDF of the form
fX(x)=ce−4x2−24x+30,
where c is a normalizing constant. Then
E[X]=-3
var(X)=1/8
Well, well, well! Looks like we have ourselves a probability density function to play with. Let's dive in, shall we?
So, we're given that fX(x) = c * e^(-4x^2 - 24x + 30). Our mission, if we choose to accept it, is to find this mysterious normalizing constant, c.
To normalize the PDF, we need to ensure that the total area under the curve is equal to 1. In other words, the sum of all probabilities must be a party animal of value 1.
Now, to do this, we're going to engage some fancy math moves. We integrate fX(x) over its entire range. That's right, we're taking a mathematical stroll from negative infinity to positive infinity.
So, let's put on our math hats and evaluate the integral of fX(x) with respect to x. After a bit of juggling, I mean, integration, we end up with:
∫{from -∞ to +∞} c * e^(-4x^2 - 24x + 30) dx
Now, my calculations might be comically off, but trust me when I say that the integral of that expression gives us a value. And to tame this infinity-loving expression, we'll call it Z.
Now, if we want the sum of all probabilities to be 1, we need the value of Z to be 1. So, our normalizing constant, c, is simply 1/Z.
That's it! We've uncovered the secret identity of the normalizing constant, c. It's 1 divided by the integral of the PDF over its entire range. But remember, this bot's math skills are a bit clownish, so double-check my answer to be on the safe side!
And as always, math with a sprinkle of humor keeps the mathematical spirits high!
To find the value of the normalizing constant c, we need to use the fact that the total area under the probability density function (PDF) is equal to 1.
The PDF is given as:
fX(x) = ce^(-4x^2-24x+30)
To find c, we need to integrate the PDF over its entire range and set it equal to 1. Thus, we have:
1 = ∫fX(x) dx
Now, let's calculate the integral:
1 = c ∫e^(-4x^2-24x+30) dx
To proceed with the integral, we can complete the square in the exponent:
1 = c ∫e^(-(4x^2+24x-30)) dx
= c ∫e^(-(4(x^2+6x)-30)) dx
= c ∫e^(-(4(x^2+6x+9-9)-30)) dx
= c ∫e^(-(4((x+3)^2-9)-30)) dx
= c e^(39) ∫e^(-(4(x+3)^2)) dx
Now, let's make a substitution:
u = x+3
du = dx
The integral simplifies to:
1 = c e^(39) ∫e^(-(4u^2)) du
Next, we can use the standard formula for integrating the Gaussian function:
∫e^(-(ax^2)) dx = √(π/a)
Applying this formula to our integral, we have:
1 = c e^(39) √(π/4)
Now, let's solve for c:
c = 1 / (e^(39) √(π/4))
Simplifying further, we get:
c = 2 / (e^(39) √π)
Therefore, the value of the normalizing constant c is 2 / (e^(39) √π).
To find the value of the normalizing constant c, we need to ensure that the probability density function (PDF) integrates to 1 over its entire range. In other words, we need to find the value of c that satisfies the condition:
∫fX(x)dx = 1
Let's calculate the integral:
∫ce^(-4x^2 - 24x + 30)dx
We can factor out the constant c:
c∫e^(-4x^2 - 24x + 30)dx
Now, to simplify the integral, we can complete the square in the exponent. To do this, we need to rewrite the expression inside the exponential function:
-4x^2 - 24x + 30 = -4(x^2 + 6x) + 30
To complete the square, we need to add and subtract a value that will make the expression inside the parentheses a perfect square. In this case, half of the coefficient of x (6) squared is 9, so we add and subtract 9:
-4(x^2 + 6x + 9 - 9) + 30
= -4((x + 3)^2 - 9) + 30
= -4(x + 3)^2 + 54 + 30
= -4(x + 3)^2 + 84
Now, we can rewrite our integral:
c∫e^(-4(x + 3)^2 + 84)dx
To simplify this further, we can pull the constant terms out of the integral:
c × e^84 × ∫e^(-4(x + 3)^2)dx
Now, we have a familiar form for the PDF of a normal distribution. The integral of the PDF of a normal distribution is equal to the square root of π divided by the square root of the coefficient of x^2 in the exponent (in this case, 4):
c × e^84 × √(π/4) × ∫e^(-4(x + 3)^2)dx
Simplifying this expression:
c × e^84 × (√π/2) × ∫e^(-4(x + 3)^2)dx
To find the value of the integral, we can use the substitution u = -4(x + 3)^2:
du/dx = -8(x + 3)
dx = du / (-8(x + 3))
Substituting these values into the integral:
c × e^84 × (√π/2) × ∫e^u × (1 / (-8(x + 3))) du
Next, we simplify:
c × e^84 × (√π/2) × (1 / 8) × ∫e^u / (x + 3) du
Now, we integrate:
c × e^84 × (√π/2) × (1 / 8) × ln |x + 3| + C
To finalize the calculation, we set the above expression equal to 1 and solve for c:
c × e^84 × (√π/2) × (1 / 8) × ln |x + 3| + C = 1
Solving for c:
c = (8 / e^84) × (2 / √π) × (1 / ln |x + 3| + C)
Note: The specific value of c cannot be determined without knowing the limits of integration or any other constraints.