Question

3. Does the family of distributions {f(x∣θ)=(3/2) ∗ (x−θ) 2 ∣θ−1

Answer 1

No, the expression provided
`\(f(x|\theta) = (3)/(2) \cdot (x-\theta)^2 \,|\theta-1\)` is not a valid probability density function because it does not integrate to 1 over the entire range. A proper probability density function must satisfy the normalization condition, which ensures that the total probability of all possible outcomes is equal to 1.

Step-by-step explanation:

For a function to be a valid probability density function, it must meet certain criteria, and one crucial criterion is the normalization condition:

`\[ \int_(-\infty)^(\infty) f(x|\theta) \,dx = 1 \]`

In the given family of distributions
`\(f(x|\theta) = (3)/(2) \cdot (x-\theta)^2 \,|\theta-1\),` the integral of
`\(f(x|\theta)\)` over its entire range does not equal 1, making it unsuitable as a probability density function. To verify this, we need to integrate
`\(f(x|\theta)\)` with respect to
`\(x\)` and evaluate the limits from
`\(-\infty\)` to
`\(+\infty\)`. If the result is not 1, then the function does not meet the necessary conditions.

In this case, the normalization condition is not satisfied, and thus, the family of distributions is not a valid probability density function.

In conclusion, while the expression may resemble a probability density function, it fails to fulfill the fundamental requirement of normalization, rendering it invalid for representing a probability distribution.