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9231 · 4.1

Continuous random variables — common mistakes

Common exam mistakes on 9231 Continuous random variables. Learn what loses marks, then practise the topic with Examiner’s Ink.

Exam tip 1

Since P(X=a) = 0 and P(X=b) = 0, remember that P(a ≤ X ≤ b), P(a < X ≤ b), P(a ≤ X < b), and P(a < X < b) are all identical for a continuous random variable. Do not waste time considering the endpoints.

Exam tip 2

When showing that f(x) ≥ 0, consider the nature of the function. For example, if f(x) = kx² for x > 0, you only need to show that the constant k is positive. If the function is more complex, a sketch or consideration of its minimum value may be required.

Exam tip 3

When constructing a piecewise CDF, be meticulous. Define F(x) = 0 for x less than the lower bound, F(x) = 1 for x greater than the upper bound, and the integral expression for x within the bounds. Check that your function is continuous at the boundaries.

Exam tip 4

A common mistake is confusing E(X²) with [E(X)]². Calculate E(X) and E(X²) as two separate, distinct steps before substituting them into the variance formula. This systematic approach reduces the chance of calculation errors.

Exam tip 5

When asked to find the CDF, remember it's a piecewise function. For the example f(x)=332x(4x)f(x) = \frac{3}{32}x(4-x) on [0,4][0, 4], the CDF is F(x)=0F(x)=0 for x<0x<0, F(x)=1F(x)=1 for x>4x>4, and F(x)=0x332t(4t)dtF(x) = \int_0^x \frac{3}{32}t(4-t) \, dt for 0x40 \le x \le 4. Don't forget to define the function for all real numbers.

Why is P(X=a) = 0 for a continuous random variable? It seems like the variable must take *some* value.

While the variable will indeed take a specific value, the probability of it being exactly that one value out of an infinite number of possibilities is infinitesimally small, and thus is defined as zero. Probability for a CRV is the area under the PDF curve over an interval. The area corresponding to a single point (an interval of width zero) is zero. We can only talk about the probability of X being in a range, e.g., P(a ≤ X ≤ b).

Can the probability density function f(x) have a value greater than 1?

Yes, it can. Unlike a probability, the value of f(x) is a measure of 'probability density', not probability itself. For example, a uniform distribution on the interval [0, 0.5] has a PDF of f(x) = 2 for 0 ≤ x ≤ 0.5. The key constraint is that the area under the f(x) curve, which represents the total probability, must integrate to 1.

When finding the CDF, F(x), why do I need to define it for values outside the main range of the PDF?

The cumulative distribution function, F(x) = P(X ≤ x), must be defined for all real numbers x from -∞ to +∞. For a PDF defined on an interval [a, b], the probability of X being less than 'a' is zero, so F(x) = 0 for x < a. The probability of X being less than or equal to any value greater than 'b' is one, as all possible values are within that range. Therefore, F(x) = 1 for x ≥ b. A complete piecewise definition is required for a full description of the distribution.