Why Translate the Event?
A probability question may describe a condition on a function of a random variable rather than directly on the variable itself. For example, it may ask for the probability that a squared measurement is no greater than a threshold, or that a nonlinear score exceeds a specified value. The distribution of \(X\) is already known; the task is to identify which values of \(X\) make the condition true.
The central step is to rewrite the event before calculating its probability. This is especially important when \(g\) is not monotone: one value of \(g(X)\) may correspond to values of \(X\) in more than one interval. A probability calculation should use all and only the values that satisfy the original condition.
For an inequality such as \(g(X)\le c\), the corresponding set is \(A=\{x:g(x)\le c\}\). Once that set is found, calculate \(P(X\in A)\) from the distribution of \(X\). No new distribution for \(g(X)\) is needed.
This equality holds because the two events describe exactly the same outcomes. In particular, the random variable \(X\) has the same value in either description; the event is simply written in two different ways. The distribution used in the probability calculation remains the distribution of \(X\).
A Reliable Translation Procedure
First solve the inequality in terms of a generic value \(x\). Do not substitute \(X\) into algebraic manipulations as though it were a fixed number; instead, determine the set of real \(x\) values for which the inequality holds. Then express the event using that set and calculate its probability under the given distribution.
For example, replace \(g(X)\le c\) with the inequality \(g(x)\le c\) to solve.
Pay attention to whether the function increases or decreases, and whether the inequality is strict.
Write the result as a condition or union of conditions on \(X\).
Sum the relevant PMF values, integrate the density over the relevant intervals, or use the CDF with the appropriate endpoints.
For a discrete variable, the probability of the translated event is the sum of the probabilities at the included values. For a continuous variable with density \(f_X\), it is the integral of \(f_X\) over the included intervals. If the CDF \(F_X\) is available, interval probabilities can also be calculated from it. Endpoint conventions matter when \(X\) has point masses: \(P(X\le a)=F_X(a)\), whereas \(P(X<a)=F_X(a-)\), the left limit of the CDF at \(a\). For a continuous distribution, changing the inclusion of a single endpoint does not change the probability, but that should not be assumed for a discrete distribution.
Worked Examples
Worked Example: A Nonmonotone Function of a Discrete Variable
A hypothetical model has a discrete random variable \(X\) with the following PMF. Find \(P((X-1)^2\le 2)\).
| \(x\) | \(-2\) | \(-1\) | \(0\) | \(1\) | \(2\) | \(3\) |
|---|---|---|---|---|---|---|
| \(P(X=x)\) | 0.10 | 0.15 | 0.20 | 0.25 | 0.20 | 0.10 |
Method. Translate the event by evaluating the inequality at each possible value of \(X\), then add the corresponding PMF values.
The event condition on a possible value \(x\) is \((x-1)^2\le 2\). Evaluate the left side at each value:
The condition holds at \(x=0,1,2\), and fails at \(x=-2,-1,3\). Therefore,
Check. The probabilities of the excluded values sum to \(0.10+0.15+0.10=0.35\), and \(0.65+0.35=1.00\). The answer is \(0.65\). This value-by-value method avoids assuming the function is one-to-one or that the event corresponds to a single interval.
Worked Example: Solving a Quadratic Inequality Under a Uniform Distribution
Suppose \(X\) has the uniform distribution on \([0,4]\), with density \(f_X(x)=1/4\) for \(0\le x\le4\). Find \(P(X(4-X)\le3)\).
Method. Solve the quadratic inequality in \(x\), then integrate the given density over the resulting intervals.
Start with the event condition:
The product is nonnegative when \(x\le1\) or \(x\ge3\). Thus the event can be written \(X\le1\) or \(X\ge3\). The density is supported on \([0,4]\), so the relevant intervals are \([0,1]\) and \([3,4]\). The probability is
The endpoints do not affect this continuous probability. As a check on the translated set, \(x=0\) satisfies the original inequality, while \(x=2\) does not because \(2(4-2)=4>3\). The answer is \(0.5000\), to four decimal places. The important algebraic step is retaining the outside intervals when the factored quadratic is nonnegative.
Worked Example: A Two-Tail Event for a Normal Variable
Let \(X\sim N(2,4)\), so \(X\) is normal with mean \(2\) and variance \(4\), and standard deviation \(2\). Find \(P((X-1)^2\ge4)\).
Method. Translate the squared inequality into two tails, then standardize each endpoint using the distribution of \(X\).
Taking square roots gives \(|X-1|\ge2\). This is equivalent to \(X-1\le-2\) or \(X-1\ge2\), hence \(X\le-1\) or \(X\ge3\). Therefore,
Here \(Z\sim N(0,1)\), and \(\Phi\) is its CDF. The two tails are disjoint, so their probabilities add. The result is \(0.3753\), to four decimal places. The calculation uses the variance \(4\) through the standard deviation \(2\); dividing by \(4\) would standardize incorrectly.
Why Monotonicity and Endpoints Matter
If \(g\) is increasing on the relevant range, an inequality involving \(g(X)\) may translate into a one-sided condition on \(X\). If \(g\) is decreasing, the direction of the inequality reverses. If \(g\) changes direction, the translated event can be a union of intervals, as in the quadratic examples. It is therefore unsafe to manipulate the inequality as though every function preserves order.
Strictness must also be carried through the translation. For example, \(g(X)<c\) and \(g(X)\le c\) need not define the same event. They differ whenever \(X\) can take a value \(x\) satisfying \(g(x)=c\). If such a value has positive probability, changing the endpoint can change the answer. With a continuous variable, a single endpoint has probability zero, but an endpoint convention can still be important for discrete or mixed distributions.
Common Exam Trap
A frequent error is to solve a quadratic inequality with the wrong intervals. For example, suppose \(X\) is uniform on \([0,5]\) and the event is \(X(5-X)\le4\). The correct algebra is
The solution is \(x\le1\) or \(x\ge4\), not \(1\le x\le4\). Restricting to the distribution's interval \([0,5]\), the correct probability is
The incorrect middle interval would give \(3/5=0.6\), which is not the probability of the stated event. A quick check helps catch the error: \(x=0\) makes the product zero and must be included, while \(x=2\) gives \(2(5-2)=6>4\) and must be excluded.
Check Your Understanding
For each problem, first rewrite the event as a condition on \(X\), then use the stated distribution to calculate its probability.
- A discrete variable \(X\) takes values \(-1,0,1,2\) with respective probabilities \(0.20,0.30,0.10,0.40\). Find \(P((X-1)^2<2)\). (Answer: \(0.80\))
- \(X\) is uniform on \([0,5]\). Find \(P((X-2)(X-4)\le0)\). (Answer: \(0.4000\))
- \(X\) has an exponential distribution with rate \(0.5\), so \(P(X>x)=e^{-0.5x}\) for \(x\ge0\). Find \(P(e^{-X}<e^{-2})\). (Answer: \(0.3679\))
- \(X\sim N(10,9)\), with mean \(10\) and variance \(9\). Find \(P((X-10)^2>9)\). (Answer: \(0.3173\))
- A discrete variable \(X\) takes values \(-2,0,1,4\) with respective probabilities \(0.10,0.40,0.30,0.20\). Find \(P(X^2-2X>0)\). (Answer: \(0.30\))