Suppose that a crazy scientist is traveling with his car in the desert. All of a sudden, he realizes that the engine's oil lamp is broken. Hence, he needs to estimate the remaining oil level to determine whether he can continue driving or should instead seek help. Since he has studied Bayesian statistics before, he wants to set up an experiment to estimate the oil level of his car, but for this, he needs a prior distribution on the oil level. Let us denote the density of the prior on the oil level by f_ o(x). The scientists knows that at any given time, the oil level x is guaranteed to be a real number in the interval [0,100].
Which properties should f_ o(\cdot ) obey in order to result in a valid probability distribution and also reflect the scientist's knowledge of the car? (Choose all that apply.)
f_ o(x)\geq 0, for every x\in [0,100].
\displaystyle \int _0^{100}f_ o(x)\; dx = 1.
\displaystyle \int _{-\infty }^{0}f_ o(x)\; dx =0.
\displaystyle \int _{100}^\infty f_ o(x)\; dx= 0.
\max _{x\in [0,100]}f_ o(x)=f_ o(50), namely, f_ o(\cdot ) attains its maximum, in the midpoint of the interval.
f_ o(\cdot ) should be a decreasing function, on [0,50), and on (50,\infty ].
The properties that the prior distribution f_ o(\cdot) should obey are:
1. f_ o(x) ≥ 0, for every x in [0,100]. This property ensures that the prior distribution is always non-negative.
2. ∫₀¹⁰⁰ f_ o(x) dx = 1. This property ensures that the total probability of all possible oil levels is equal to 1, representing a valid probability distribution.
3. ∫₋∞⁰ f_ o(x) dx = 0. Since the oil level cannot be negative, the probability of having a negative oil level is zero.
4. ∫₁₀⁰⁺ f_ o(x) dx = 0. Similarly, since the oil level cannot exceed 100, the probability of having an oil level greater than 100 is zero.
5. max_{x∈[0,100]} f_ o(x) = f_ o(50), namely, f_ o(\cdot) attains its maximum at the midpoint of the interval. This property reflects the scientist's initial knowledge that there is no specific reason to believe the oil level is more likely to be at one end of the interval than the other.
6. f_ o(\cdot) should be a decreasing function on [0,50) and on (50,∞). This property reflects the scientist's initial belief that the oil level is more likely to be lower than higher, given that the car has been driven for some time.
So, the correct properties that the prior distribution should have are:
- f_ o(x) ≥ 0, for every x in [0,100].
- ∫₀¹⁰⁰ f_ o(x) dx = 1.
- ∫₋∞⁰ f_ o(x) dx = 0.
- ∫₁₀⁰⁺ f_ o(x) dx = 0.
- max_{x∈[0,100]} f_ o(x) = f_ o(50), namely, f_ o(\cdot) attains its maximum at the midpoint of the interval.
- f_ o(\cdot) should be a decreasing function on [0,50) and on (50,∞).
The properties that the prior distribution should obey in order to result in a valid probability distribution and reflect the scientist's knowledge of the car are:
1. f_ o(x) ≥ 0, for every x ∈ [0,100].
This property ensures that the probability density function is non-negative for all values of x within the specified range.
2. ∫₀¹₀₀ f_ o(x) dx = 1.
This property ensures that the total probability over the entire interval [0,100] is equal to 1, reflecting the fact that the oil level must be within this range.
3. ∫₋₋∞₀ f_ o(x) dx = 0.
This property ensures that the probability of the oil level being less than 0 is zero, as negative oil levels are not physically possible.
4. ∫₁₀₀₋∞ f_ o(x) dx = 0.
This property ensures that the probability of the oil level being greater than 100 is zero, as oil levels exceeding 100 are not physically possible.
5. maxₓ∈[0,100] f_ o(x) = f_ o(50), namely, f_ o(·) attains its maximum in the midpoint of the interval.
This property reflects the scientist's belief that the most likely oil level is at the midpoint of the interval, assuming no other information is known.
6. f_ o(·) should be a decreasing function on [0,50) and on (50,∞).
This property reflects the scientist's knowledge that as the oil level increases beyond 50, it becomes less likely due to usage and potential leakage.
Therefore, the correct properties that the prior distribution should obey are:
- f_ o(x) ≥ 0, for every x ∈ [0,100].
- ∫₀¹₀₀ f_ o(x) dx = 1.
- ∫₋₋∞₀ f_ o(x) dx = 0.
- ∫₁₀₀₋∞ f_ o(x) dx = 0.
- maxₓ∈[0,100] f_ o(x) = f_ o(50), namely, f_ o(·) attains its maximum in the midpoint of the interval.
- f_ o(·) should be a decreasing function on [0,50) and on (50,∞).