ML Interview Q Series: Closest Odd Integer Probability for Exponential Variables: The Role of Memorylessness.
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You simulate a random observation from an exponentially distributed random variable X with expected value 1. What is the probability that the closest integer to the random observation is odd? What is the probability if the random observation is larger than a given even integer r? Can you explain why the two probabilities are the same?
Short Compact solution
The probability that the closest integer to X is odd turns out to be
When X is conditioned on being larger than an even integer r, the probability that the closest integer is odd remains the same, namely
This follows directly from the memoryless property of the exponential distribution, which ensures that the behavior of X after exceeding r is statistically identical (up to a shift) to the original distribution, thus preserving the same probability of rounding to an odd integer.
Comprehensive Explanation
To see why the unconditional probability is e^(-1/2)/(1 + e^(-1)), consider how to determine if X is closer to an odd integer or to an even integer. For X > 0, the point x belongs to an interval that is closest to an odd integer if x lies in half-integer “bands” around each odd integer. Specifically, for k = 0, 1, 2, …, the odd integer of interest is 2k+1, and the half-integer band around it is (2k + 1/2, 2k + 1 + 1/2). By summing probabilities over all these disjoint bands, we obtain the overall probability that the rounded value is odd.
Let f(x) = e^(-x) for x >= 0 be the probability density function of X, and F(x) = 1 - e^(-x) be its cumulative distribution function. Then the probability that X falls in one band (2k + 1/2, 2k + 1 + 1/2) is e^(-(2k + 1/2)) - e^(-(2k + 1 + 1/2)), because each band’s probability is F(2k+1+1/2) - F(2k+1/2). Summing over k from 0 to infinity leads to a telescoping series that simplifies to e^(-1/2)/(1 + e^(-1)).
Conditioning on X > r, where r is an even integer, does not alter that probability. The exponential distribution has the memoryless property:
Hence, once you know X > r, the distribution of (X - r) is still exponential with the same rate. The “pattern” of half-integer intervals beyond r is effectively the same as it was from 0 onward, just shifted by r. Because r is even, the positions of the odd integers beyond r follow the same half-integer offset that we see from 0 onward. This yields the same probability of rounding to an odd integer after r as the unconditional case.
Follow-up Questions
How exactly does the memoryless property ensure the same probability?
The memoryless property states that for any s and t > 0, P(X > s + t | X > s) = P(X > t). This means that once we condition on surviving past s, the “starting point” effectively resets, and X behaves as though it has just begun an exponential clock. In our context, setting s = r (an even integer) allows us to assert that the distribution of X - r, given X > r, is the same as the original distribution of X. Since the intervals around odd integers beyond r mirror the intervals around odd integers beyond 0, the probability remains unchanged.
What if r were odd? Would the probability be the same?
If r were odd, we would then shift the random variable by that odd integer. The new intervals that define “closest integer is odd” versus “closest integer is even” would again retain the same structure (just offset by an odd integer). Because the exponential is memoryless, the probability of landing in an “odd-integer band” beyond r would still match the original probability. In essence, the argument does not require r to be even; the key requirement is that the exponential distribution’s shift-invariance holds for any r.
What if the random variable did not have the memoryless property?
If X had a different distribution (for instance, a Gamma distribution with shape parameter > 1 or a bounded distribution), conditioning on X > r could change the tail behavior, thereby altering how likely we are to land near an odd integer. In that case, the probability that the closest integer is odd would not necessarily remain the same under the condition X > r. The distinctive feature of the exponential distribution is exactly that it lacks “aging,” so the relative placement in the half-integer intervals does not change, no matter how far along X has already gone.
How can we generalize this to continuous-time Markov processes?
In a continuous-time Markov chain with certain “renewal” or “reset” properties, you can see analogous ideas where, after a certain state, the process restarts with the same distribution over subsequent paths. Any time-homogeneous Markov chain with the property that “starting fresh” from a state replicates the behavior from time 0 can exhibit similar phenomena. The exponential distribution is often used to model waiting times between state transitions exactly because of this memoryless feature.
Could we verify this probability numerically?
Yes. One simple approach is to simulate many exponential random variables with mean 1, compute whether they round to an odd or even integer, and track the empirical fraction of odd rounding. In Python:
import numpy as np
n = 10_000_000
X = np.random.exponential(scale=1.0, size=n) # mean = 1
rounded = np.round(X).astype(int)
p_odd = np.mean(rounded % 2 == 1)
print("Empirical Probability of Odd:", p_odd)
As n grows large, p_odd will approach e^(-0.5)/(1 + e^(-1)).
To check the conditional probability, we can filter out samples where X <= r (with r even) and then recalculate the fraction of odd rounding among the remaining samples:
r = 10 # example even integer
larger_than_r = X[X > r]
rounded_cond = np.round(larger_than_r).astype(int)
p_odd_cond = np.mean(rounded_cond % 2 == 1)
print("Empirical Conditional Probability of Odd Given X > r:", p_odd_cond)
We will observe that this probability remains very close to the same value, matching the theoretical result from the memoryless property.
Below are additional follow-up questions
Can the probability be interpreted via geometric considerations on the real line?
Yes, by considering how an exponential distribution decays on the positive real line. One might interpret the question of “closest integer being odd” as a geometric partition of the real axis into half-integer neighborhoods around each integer. Specifically, each integer m has a neighborhood (m – 0.5, m + 0.5). The question becomes: what fraction of the mass of the exponential distribution falls into neighborhoods around odd integers? Because the exponential distribution is monotonically decreasing, one might suspect that these neighborhoods near zero (especially those closer to zero) hold higher weight. However, the exact fraction is determined by a telescoping sum of exponentials, and the memoryless property ensures that no matter how far out you go, the proportion of intervals capturing “odd integers” remains the same. This interpretation highlights the important interplay between geometry (interval partitioning) and distributional properties (exponential decay and memorylessness).
What happens if we consider a continuous shift that is not necessarily an integer?
If we shift X by some arbitrary amount c > 0 that is not necessarily an integer, we are effectively looking at P(closest integer to (X + c) is odd). This shift changes where the “odd integer neighborhoods” land relative to X. But the memoryless property still means that X conditioned on X > r behaves like X + constant for some distribution. The subtlety here is that an integer shift that aligns with the pattern of odd/even integers preserves the same partition structure. With a non-integer shift, the locations of the half-integer cuts shift in a non-integer manner, and the fraction of mass in the shifted intervals might not remain the same. So the probability that the “closest integer is odd” for X + c can differ from the unshifted scenario—unless c is an integer. This underscores that the memorylessness preserves distribution shape but does not necessarily align integer boundaries the same way if the shift is not integral.
How would the result change if the exponential distribution had a different rate parameter?
For an exponential distribution with rate parameter λ (meaning E[X] = 1/λ), the PDF is f(x) = λ exp(-λ x) for x >= 0. In that setting, the question becomes: “What is P(closest integer to X is odd) for X ~ Exp(λ)?” We can still sum over intervals (2k + 1/2, 2k + 1 + 1/2), but the probabilities would use e^(-λ x) terms. The summation becomes sum_{k=0 to infinity} [e^(-λ(2k + 1/2)) - e^(-λ(2k + 1 + 1/2))]. Because the memoryless property still holds for exponentials with any positive λ, the same fraction would apply conditionally on X > r, but the constant in the final expression changes from e^(-1/2) to e^(-λ/2), and from e^(-1) to e^(-λ). The logic remains identical, but the numeric value changes with λ. This ensures we understand that changing the mean simply rescales the distribution in a way that proportionally affects each exponential term, but does not break memorylessness or the telescoping series argument.
How would one handle a discrete version of this problem with a geometric distribution?
In a discrete setting, say a random variable Y that follows a geometric distribution (with P(Y = k) = p (1 – p)^k for k = 0, 1, 2, …), the question of “closest integer is odd or even” becomes somewhat trivial because Y itself is already integer-valued. So, the probability that the “closest integer is odd” is the probability that Y is odd, i.e. sum_{k odd} p (1 – p)^k. This might be a simpler summation but loses the nuance of half-integer cutoff points. Nonetheless, memorylessness still holds in the discrete sense (geometric distribution is also memoryless), and the conditional probability that Y is odd given Y > r remains the same as the unconditional probability of being odd. This helps us see that memorylessness in both discrete and continuous domains preserves the distribution over “future outcomes,” thus answering a related but distinct scenario.
Could any boundary effects appear if we only observe X above some threshold close to 0?
Yes. If the threshold r is close to 0 (especially less than 1), we might worry about intervals that cross 0 or about the possibility that X cannot be below 0. However, the exponential distribution is 0 on the negative side by definition, so we only care about the region X >= 0. When r is a positive even integer, any boundary effects near 0 do not matter because we are conditioning on X > r, and r >= 2 in that sense. The memorylessness ensures that once X surpasses r, the “time spent” below r is irrelevant. An edge case might be r = 0 exactly, but that simply reduces to the unconditional scenario. Hence, boundary effects near 0 have no real impact in the standard problem as long as r > 0 is well-defined.
If we had a truncated exponential distribution, would the memoryless property still hold?
No. The classic exponential distribution is supported on [0, ∞). If we artificially truncate the variable, for instance at some maximum M, the resulting truncated distribution is not memoryless. In that case, conditioning on X > r changes the conditional distribution in a more complex way because we also know X <= M. This can shift the odds of hitting intervals near odd integers. As a result, the probability that the closest integer is odd, conditional on X > r, might differ from the unconditional probability. This is a subtle but important point: memorylessness only strictly holds for an unbounded exponential on [0, ∞). Truncating it breaks that property.
How might rounding conventions (e.g., ties to even) affect the result?
The question states “closest integer,” but in typical rounding rules, if X = n + 0.5 exactly, one might choose to round to the even integer. This is known as “round half to even.” If we incorporate that nuance, then the intervals that define rounding to an odd integer would exclude the boundary points where X = integer + 0.5 if that boundary integer is even. The effect on the probability can be extremely small because the probability that X hits an exact half-integer is 0 in continuous distributions. Strictly speaking, from a measure-theoretic standpoint, the set of half-integers has measure zero, so it does not affect the probability. However, in a real-world implementation with floating-point arithmetic, small numerical artifacts could appear. For the theoretical analysis, these boundary points do not contribute, so the result remains the same.
Does the same probability hold if we consider “farthest integer” or some other unusual measure of closeness?
No, the problem specifically asks about the “closest integer,” which is a natural partition of the real line into 1-unit segments centered on each integer. If we define “farthest integer” or any other non-standard measure, we would have to redefine the sets in which X falls. For example, “farthest integer from X is odd” would be a more unusual partition that does not match straightforward intervals of length 1. The exponential distribution is decreasing, but not symmetric, so such a definition might lead to unintuitive or different partition structures. The memoryless property alone would not guarantee the same fraction for “farthest integer is odd,” because these intervals around integers would likely overlap in more complicated ways. Thus the result and proof technique do not automatically generalize to such a scenario.
Could discretization errors in a real system simulation significantly affect the probability estimate?
In practical numeric simulations, floating-point arithmetic has finite precision. If we simulate very large or very small values of X, numerical underflow or overflow might occur when computing e^(-x). This can skew probabilities if not handled carefully. Some specific pitfalls are:
Very large x can lead to e^(-x) underflowing to 0, which might cause the tail probabilities to appear smaller than they truly are.
Very small differences might be lost in floating-point representation.
Summation of large numbers of tiny probabilities might accumulate error. In typical double-precision scenarios, these effects are usually minor for moderate ranges. But for extreme values (e.g., r = 10^6), one might need arbitrary-precision arithmetic or carefully designed approximations. While these are implementation details, they can cause an empirical probability to deviate from the theoretical e^(-1/2) / (1 + e^(-1)) if not carefully managed.