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Sagot :
To verify whether [tex]\( f(x) = \frac{1}{3} \)[/tex] for [tex]\( 0 \leq x \leq 3 \)[/tex] and [tex]\( f(x) = 0 \)[/tex] otherwise is a probability density function (pdf), we need to check two essential conditions:
1. Non-negativity: The function [tex]\( f(x) \)[/tex] must be non-negative for all values of [tex]\( x \)[/tex].
2. Normalization: The total integral of [tex]\( f(x) \)[/tex] over the entire range must be equal to 1.
### Step 1: Non-negativity
We need to ensure [tex]\( f(x) \geq 0 \)[/tex] for all [tex]\( x \)[/tex].
- For [tex]\( 0 \leq x \leq 3 \)[/tex], [tex]\( f(x) = \frac{1}{3} \)[/tex]. Since [tex]\(\frac{1}{3} \)[/tex] is a positive number, [tex]\( f(x) \geq 0 \)[/tex] in this interval.
- For [tex]\( x < 0 \)[/tex] or [tex]\( x > 3 \)[/tex], [tex]\( f(x) = 0 \)[/tex] which is also non-negative.
Thus, [tex]\( f(x) \geq 0 \)[/tex] for all [tex]\( x \)[/tex].
### Step 2: Normalization
We need to check if the total integral of [tex]\( f(x) \)[/tex] over the entire range equals 1.
[tex]\[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \][/tex]
Splitting the integral into the defined ranges of [tex]\( f(x) \)[/tex], we have:
[tex]\[ \int_{-\infty}^{\infty} f(x) \, dx = \int_{-\infty}^{0} f(x) \, dx + \int_{0}^{3} f(x) \, dx + \int_{3}^{\infty} f(x) \, dx \][/tex]
Since [tex]\( f(x) = 0 \)[/tex] for [tex]\( x < 0 \)[/tex] and [tex]\( x > 3 \)[/tex], these integrals will be zero:
[tex]\[ \int_{-\infty}^{0} f(x) \, dx = 0 \][/tex]
[tex]\[ \int_{3}^{\infty} f(x) \, dx = 0 \][/tex]
So we are left with:
[tex]\[ \int_{-\infty}^{\infty} f(x) \, dx = \int_{0}^{3} f(x) \, dx \][/tex]
For [tex]\( 0 \leq x \leq 3 \)[/tex], [tex]\( f(x) = \frac{1}{3} \)[/tex], thus:
[tex]\[ \int_{0}^{3} f(x) \, dx = \int_{0}^{3} \frac{1}{3} \, dx \][/tex]
Evaluating this integral:
[tex]\[ \int_{0}^{3} \frac{1}{3} \, dx = \frac{1}{3} \times (3 - 0) = \frac{1}{3} \times 3 = 1 \][/tex]
Since the integral over the entire range equals 1, the normalization condition is satisfied.
### Conclusion
Since [tex]\( f(x) \)[/tex] is non-negative for all [tex]\( x \)[/tex] and its integral over the entire range is equal to 1, [tex]\( f(x) \)[/tex] satisfies the conditions to be a probability density function.
Therefore, [tex]\( f(x) = \frac{1}{3} \)[/tex] for [tex]\( 0 \leq x \leq 3 \)[/tex] and [tex]\( f(x) = 0 \)[/tex] otherwise is indeed a probability density function. The validation shows that the total area under the curve [tex]\( f(x) \)[/tex] is 1, confirming that it can represent a valid pdf.
1. Non-negativity: The function [tex]\( f(x) \)[/tex] must be non-negative for all values of [tex]\( x \)[/tex].
2. Normalization: The total integral of [tex]\( f(x) \)[/tex] over the entire range must be equal to 1.
### Step 1: Non-negativity
We need to ensure [tex]\( f(x) \geq 0 \)[/tex] for all [tex]\( x \)[/tex].
- For [tex]\( 0 \leq x \leq 3 \)[/tex], [tex]\( f(x) = \frac{1}{3} \)[/tex]. Since [tex]\(\frac{1}{3} \)[/tex] is a positive number, [tex]\( f(x) \geq 0 \)[/tex] in this interval.
- For [tex]\( x < 0 \)[/tex] or [tex]\( x > 3 \)[/tex], [tex]\( f(x) = 0 \)[/tex] which is also non-negative.
Thus, [tex]\( f(x) \geq 0 \)[/tex] for all [tex]\( x \)[/tex].
### Step 2: Normalization
We need to check if the total integral of [tex]\( f(x) \)[/tex] over the entire range equals 1.
[tex]\[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \][/tex]
Splitting the integral into the defined ranges of [tex]\( f(x) \)[/tex], we have:
[tex]\[ \int_{-\infty}^{\infty} f(x) \, dx = \int_{-\infty}^{0} f(x) \, dx + \int_{0}^{3} f(x) \, dx + \int_{3}^{\infty} f(x) \, dx \][/tex]
Since [tex]\( f(x) = 0 \)[/tex] for [tex]\( x < 0 \)[/tex] and [tex]\( x > 3 \)[/tex], these integrals will be zero:
[tex]\[ \int_{-\infty}^{0} f(x) \, dx = 0 \][/tex]
[tex]\[ \int_{3}^{\infty} f(x) \, dx = 0 \][/tex]
So we are left with:
[tex]\[ \int_{-\infty}^{\infty} f(x) \, dx = \int_{0}^{3} f(x) \, dx \][/tex]
For [tex]\( 0 \leq x \leq 3 \)[/tex], [tex]\( f(x) = \frac{1}{3} \)[/tex], thus:
[tex]\[ \int_{0}^{3} f(x) \, dx = \int_{0}^{3} \frac{1}{3} \, dx \][/tex]
Evaluating this integral:
[tex]\[ \int_{0}^{3} \frac{1}{3} \, dx = \frac{1}{3} \times (3 - 0) = \frac{1}{3} \times 3 = 1 \][/tex]
Since the integral over the entire range equals 1, the normalization condition is satisfied.
### Conclusion
Since [tex]\( f(x) \)[/tex] is non-negative for all [tex]\( x \)[/tex] and its integral over the entire range is equal to 1, [tex]\( f(x) \)[/tex] satisfies the conditions to be a probability density function.
Therefore, [tex]\( f(x) = \frac{1}{3} \)[/tex] for [tex]\( 0 \leq x \leq 3 \)[/tex] and [tex]\( f(x) = 0 \)[/tex] otherwise is indeed a probability density function. The validation shows that the total area under the curve [tex]\( f(x) \)[/tex] is 1, confirming that it can represent a valid pdf.
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