Why These Distinctions Matter
Many probability errors arise not from difficult calculations but from reading a distribution incorrectly. A density height can be mistaken for a probability, a formula can be used outside the random variable’s support, or an endpoint can be included when the event excludes it. These distinctions matter whenever a probability is read from a PMF, density, or CDF.
The key is to ask what each object represents. A PMF gives probabilities at individual values. A density gives probability through area over an interval. A CDF gives the probability of being at or below a specified value. These descriptions are simple, but their consequences differ at points, outside the support, and at jumps.
For a discrete variable, a PMF value is itself a probability and must be between 0 and 1. For a density, the height is not a point probability: it can even exceed 1, provided the total area under the density is 1. For a CDF, the value at \(x\) includes all probability at \(x\), because its event is \(X\le x\), not \(X<x\).
Density Height, Area, and Support
A density describes how probability is distributed across a continuous range. To find a probability over an interval, integrate the density over that interval. A density height such as \(f_X(1.5)\) is not \(P(X=1.5)\); for a continuous random variable, the probability at any single point is zero. This follows because a single point has interval width zero, so the area over that point is zero.
The support matters just as much. If a density is specified only on \([a,b]\), the probability outside \([a,b]\) is zero. A formula for the density should be read together with its stated range, not extended to every real number.
Worked Example: Density Height Is Not an Interval Probability
Suppose a continuous random variable \(X\) has density \(f_X(x)=x/2\) for \(0\le x\le 2\), and \(f_X(x)=0\) otherwise. Find \(f_X(1.5)\), \(P(0.5<X\le1.5)\), \(P(X=1)\), and \(P(X>2)\). The model is a continuous distribution supported on \([0,2]\).
Step 1: Check the density and support. The density is nonnegative on \([0,2]\). Its total area is
Thus it is a valid density, and values outside \([0,2]\) have probability zero.
Step 2: Evaluate the density height.
This is a density height, not a probability. In particular, it does not mean that \(P(X=1.5)=0.75\).
Step 3: Find the interval probability by area.
Step 4: Handle a point and the support boundary. Since \(X\) is continuous, \(P(X=1)=0\). Since \(2\) is the largest value in the support, \(P(X>2)=0\).
Check and trap avoided. The interval probability \(0.50\) lies between 0 and 1, and the density integrates to 1. The common error is to report \(0.75\) as the probability at \(1.5\); the correction is that a density height is not a point probability.
Reading CDF Values, Jumps, and Endpoints
The CDF always uses a closed upper endpoint: \(F_X(x)=P(X\le x)\). Consequently, a point mass at \(x\) is included in \(F_X(x)\). The CDF just to the left of \(x\), written \(F_X(x-)\), excludes any mass at \(x\). Their difference is the probability of that point.
For interval events, subtract the CDF values that match the event’s endpoints. For example, \(P(a<X\le b)=F_X(b)-F_X(a)\): the event \(X\le a\) is removed from the event \(X\le b\), leaving values strictly above \(a\) and at most \(b\). Other endpoint forms follow by including or removing the mass at the relevant endpoint.
For a continuous distribution, there are no point masses, so \(F_X(x)=F_X(x-)\) at every point and changing an interval endpoint does not change its probability. For a discrete or mixed distribution, that shortcut is not valid at an atom (a point with positive probability).
Worked Example: A Mixed Distribution with Endpoint Masses
Let \(X\) have probability \(0.2\) at \(0\), density \(0.2\) on \(0<x<3\), and probability \(0.2\) at \(3\). There is no probability elsewhere. Find the CDF, \(P(X=0)\), \(P(X=3)\), and \(P(0<X\le3)\). This is a mixed distribution: it has two point masses and a continuous component.
Step 1: Verify the total probability. The continuous component has area \(3(0.2)=0.6\). Adding the two point masses gives \(0.2+0.6+0.2=1\).
Step 2: Build the CDF from the probability accumulated so far. Below 0, the accumulated probability is 0. At 0, the CDF jumps to \(0.2\). Between 0 and 3, the continuous component adds area \(0.2x\) to that initial mass. Just before 3, the value is \(0.2+0.2(3)=0.8\). At 3, the final mass raises it to 1. Thus
Step 3: Read the endpoint masses from jumps. At 0, \(F_X(0)=0.2\) and \(F_X(0-)=0\), so \(P(X=0)=0.2-0=0.2\). At 3, \(F_X(3)=1\) and \(F_X(3-)=0.8\), so \(P(X=3)=1-0.8=0.2\).
Step 4: Calculate the interval probability with its stated endpoints. The event \(0<X\le3\) excludes the mass at 0 and includes the mass at 3. Therefore,
This also equals the continuous probability \(0.6\) plus the mass at 3, \(0.2\). The answer is \(0.8\), not \(0.6\): overlooking the included endpoint mass is the common mistake. The CDF calculation and the component-by-component check agree.
PMFs and the Support of a Discrete Variable
A PMF assigns probability only to values the variable can take. If \(X\) can equal only a finite set of values, then all other points have probability zero, even if they lie between values in that set. The probabilities in the PMF must be nonnegative and sum to 1. Its CDF is a step function: it increases by the PMF value at each supported point.
Worked Example: Support and the CDF of a Discrete Variable
A hypothetical inspection score \(X\) takes values \(-1\), \(2\), and \(5\), with probabilities \(0.25\), \(0.50\), and \(0.25\), respectively. Find \(F_X(2-)\), \(F_X(2)\), \(P(X<2)\), \(P(X\le2)\), and \(P(2<X\le5)\). No other values are in the support.
Step 1: Check the PMF. The probabilities are nonnegative and sum to \(0.25+0.50+0.25=1\).
Step 2: Accumulate only the masses at supported values. Just to the left of 2, only the mass at \(-1\) has been included, so \(F_X(2-)=0.25\). At 2, the mass \(0.50\) is included:
Step 3: Match each event to its endpoint rule. Since the support has no value between \(-1\) and 2, \(P(X<2)=0.25\). The closed event \(X\le2\) includes the mass at 2, so \(P(X\le2)=0.75\). Finally, \(2<X\le5\) contains only the value 5, giving \(P(2<X\le5)=0.25\).
Check and trap avoided. The jump at 2 is \(0.75-0.25=0.50\), exactly the PMF value at 2. The common error is to treat \(F_X(2)\) as \(P(X<2)\); the CDF convention includes equality.
A Practical Error-Checking Routine
When a probability calculation looks suspicious, identify the object and the event before doing arithmetic. The following checks address the most frequent misreadings.
A PMF value is a point probability, a density value is a height, and a CDF value is the probability \(P(X\le x)\).
Do not apply a formula outside the values the variable can take. Events entirely outside the support have probability zero.
For a CDF, \(F_X(x)\) includes a mass at \(x\), while \(F_X(x-)\) excludes it. At a continuous point, the jump is zero.
A probability must lie between 0 and 1. PMF values must sum to 1, and a density must have total area 1 over its support.
A density value greater than 1 is not automatically an error: density is measured per unit of the variable, while probability is area. Conversely, a PMF entry of 1.2 is impossible because a point probability cannot exceed 1. Keep those checks distinct.
Check Your Understanding
For each problem, identify the relevant support or endpoint convention before calculating.
- A continuous random variable has density \(f_X(x)=1/4\) for \(2\le x\le6\), and zero otherwise. Find \(P(X=4)\), \(P(3<X\le5)\), and \(P(X<2)\). (Answer: \(0,\ 0.5000,\ 0\))
- A discrete variable has support \(\{-2,1,4\}\) with respective probabilities \(0.3,0.4,0.3\). Find \(P(X<1)\), \(P(X\le1)\), and \(P(X=1)\). (Answer: \(0.3000,\ 0.7000,\ 0.4000\))
- A mixed variable has mass \(0.25\) at \(-1\), density \(0.25\) on \(0<x<2\), and mass \(0.25\) at 3, with no other probability. Find \(F_X(0)\), \(F_X(2)\), \(P(-1<X\le2)\), and \(P(X=3)\). (Answer: \(0.2500,\ 0.7500,\ 0.5000,\ 0.2500\))
- A proposed CDF is \(G(x)=0\) for \(x<0\), \(G(x)=0.4\) for \(0\le x<1\), \(G(x)=0.3\) for \(1\le x<2\), and \(G(x)=1\) for \(x\ge2\). Is this a valid CDF? Identify the property that decides. (Answer: No; it decreases from \(0.4\) to \(0.3\) at 1, but a CDF must be nondecreasing.)
- A continuous variable has CDF \(F_X(x)=0\) for \(x<1\), \(F_X(x)=(x-1)/4\) for \(1\le x<5\), and \(F_X(x)=1\) for \(x\ge5\). Find \(P(X=1)\) and \(P(2<X\le4)\). (Answer: \(0,\ 0.5000\))