Suppose that a random variable Y has a probability density function given by
f(y) =
(
ky
3
e
y=2
; y > 0
0; elsewhere:
Find the value of k that makes f(y) a density function.
To find the value of k that makes f(y) a density function, we need to make sure that the total area under the probability density function (PDF) is equal to 1.
In this case, the PDF is given by:
f(y) = ky^3 * e^(y/2) ; y > 0
f(y) = 0 ; elsewhere
To find the value of k, we need to calculate the integral of f(y) over its entire range.
∫[0,∞] f(y) dy = 1
∫[0,∞] k * y^3 * e^(y/2) dy = 1
Let's solve this integral step by step.
Step 1: Identify the appropriate substitution.
Let u = y/2
Therefore, y = 2u and dy = 2du
Step 2: Change the limits of integration.
When y = 0, u = 0
When y = ∞, u = ∞
Step 3: Rewrite the integral with the substitution.
∫[0,∞] k * (2u)^3 * e^u * (2du)
Step 4: Simplify the expression and calculate the integral.
2^3 = 8, so the integral becomes:
∫[0,∞] 8k * u^3 * e^u du
Using integration by parts, we can solve this integral. The formula for integration by parts is:
∫ u * v dx = u * ∫ v dx - ∫ (u' * ∫ v dx) dx
Let's apply integration by parts to our integral.
Step 5: Identify u and dv.
Let u = u^3
dv = 8k * e^u du
Step 6: Calculate du and v.
du = 3u^2 du
v = ∫ dv = 8k * e^u du = 8k * e^u
Step 7: Apply the integration by parts formula.
Using the formula, we have:
∫ u * v dx = u * ∫ v dx - ∫ (u' * ∫ v dx) dx
∫ u * dv = u * v - ∫ v * du
Plugging in the values we identified earlier, we get:
∫ (u^3 * 8k * e^u du) = (u^3 * 8k * e^u) - ∫ (3u^2 * 8k * e^u du)
Simplifying this expression, we have:
∫ 8k * u^3 * e^u du = 8k * u^3 * e^u - 24k * ∫ u^2 * e^u du
Step 8: Continue applying integration by parts.
Using integration by parts again, we have:
∫ u^2 * e^u du = u^2 * e^u - ∫ 2u * e^u du
Simplifying this expression, we have:
∫ u^2 * e^u du = u^2 * e^u - 2 * ∫ u * e^u du
Step 9: Apply integration by parts one more time.
Using integration by parts again, we have:
∫ u * e^u du = u * e^u - ∫ e^u du
Simplifying this expression, we have:
∫ u * e^u du = u * e^u - ∫ e^u du = u * e^u - e^u
Step 10: Substitute the results back into the original integral.
Plugging the results back into the original integral:
∫ 8k * u^3 * e^u du = 8k * u^3 * e^u - 24k * (u^2 * e^u - 2 * (u * e^u - e^u))
Expanding and simplifying:
∫ 8k * u^3 * e^u du = 8k * u^3 * e^u - 24k * u^2 * e^u + 48k * u * e^u - 48k * e^u
Step 11: Evaluate the integral over its limits.
We need to calculate this integral from 0 to ∞, so we substitute the limits:
∫[0,∞] 8k * u^3 * e^u du = [8k * u^3 * e^u - 24k * u^2 * e^u + 48k * u * e^u - 48k * e^u] evaluated from 0 to ∞
As u approaches ∞, all terms multiplied by u will dominate, making them approaches infinity. Similarly, as u approaches ∞, the exponential term e^u will also approach infinity.
Therefore, we have:
[8k * u^3 * e^u - 24k * u^2 * e^u + 48k * u * e^u - 48k * e^u] evaluated from 0 to ∞ = ∞
This means that the integral diverges and does not converge to a finite value.
To make the probability density function a valid density function, the integral of f(y) over its entire range must equal 1. However, in this case, the integral diverges, which means there is no value of k that can make f(y) a density function.