ML Interview Q Series: Calculating P(X is even) using Probability Generating Functions at z=-1.
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Short Compact solution
Denote by pn the probability P(X = n). By definition of the generating function, GX(z) = Σ pn zn (summing n from 0 to ∞). If we substitute z = -1, we get GX(-1) = Σ pn (-1)n = Σ p2n - Σ p2n+1. Because Σ p2n + Σ p2n+1 = 1, it follows that GX(-1) + 1 = 2 Σ p2n. Hence, Σ p2n = ½ [GX(-1) + 1], which proves that the probability that X is even equals ½ (GX(-1) + 1).
Comprehensive Explanation
Role of the Probability Generating Function (PGF)
The probability generating function for a discrete random variable X that takes nonnegative integer values is defined as GX(z) = Σ pn zn, where pn is the probability P(X = n). This function transforms the sequence of probabilities {pn} into a function in the complex variable z. A key property is that, by strategically substituting specific values for z, we can isolate certain sums of the probability mass function.
Why Evaluate at z = -1?
When we set z = -1 in GX(z), we are effectively alternating the sign of consecutive probabilities: GX(-1) = p0 - p1 + p2 - p3 + … This alternating sum neatly separates the even-indexed probabilities from the odd-indexed probabilities: GX(-1) = (p0 + p2 + …) - (p1 + p3 + …).
Relating Even and Odd Terms to Probability 1
We know that p0 + p1 + p2 + … = 1. If we denote the sum of even-indexed probabilities by Seven = p0 + p2 + p4 + …, and the sum of odd-indexed probabilities by Sodd = p1 + p3 + p5 + …, then Seven + Sodd = 1. From the expression for GX(-1), GX(-1) = Seven − Sodd.
Hence, if GX(-1) = Seven − Sodd and we also have Seven + Sodd = 1, then solving these two equations simultaneously yields Seven = ½ [GX(-1) + 1].
Since P(X is even) = Seven, we immediately obtain P(X is even) = ½ [GX(-1) + 1].
Implementation Detail
In a practical setting, if you have a probability mass function pn, you can compute the partial sums of p2n and p2n+1 directly to verify this relationship numerically. Alternatively, if you have a closed-form expression for GX(z), you can substitute z = -1 into that expression to find GX(-1) and then use the formula.
Below is a short Python snippet illustrating how you might verify P(X is even) for a finite distribution pn:
p = [0.1, 0.2, 0.05, 0.15, 0.5] # Example PMF for X = 0..4 (sum is 1)
G_at_neg1 = sum(p[n] * ((-1)**n) for n in range(len(p)))
S_even = sum(p[n] for n in range(0, len(p), 2)) # Summation of even indices
S_from_G = 0.5 * (G_at_neg1 + 1)
print("Direct sum of even indices =", S_even)
print("From G_X(-1) relationship =", S_from_G)
How We Might Derive This More Formally
One can see the result as leveraging the identity p2n + p2n+1 = 1 - … or simply using the idea of splitting any infinite series into alternating sums. The key is that evaluating the generating function at z = -1 results in a telescoping that isolates even and odd terms with positive and negative signs, respectively.
Follow-Up Question: Could We Use Other Generating Functions?
There are closely related transforms such as the characteristic function E[eitX] or the moment generating function E[etX]. In principle, you might attempt to evaluate these at specific values of t to isolate sums of probabilities. However, for discrete random variables with nonnegative integer support, the probability generating function GX(z) is the most direct tool for capturing sums of the form Σ pn zn.
Follow-Up Question: What Happens If X Has Infinite Support?
If X can take on all nonnegative integer values, the same derivation applies, provided the probability generating function converges at z = -1. Most standard distributions like the Poisson or the Binomial (even extended to infinite support) still produce well-defined sums for z = -1. In any case, you want to ensure GX(z) is defined there.
Follow-Up Question: Edge Cases and Potential Pitfalls
A subtlety arises if the series for GX(z) diverges at z = -1. This can happen for certain long-tailed distributions where the sum Σ pn (-1)n does not converge absolutely. In standard practice, though, most well-known distributions with finite or exponentially decaying tails will converge properly at z = -1. If it converges, then the derivation using GX(-1) is valid, and the formula P(X is even) = ½ [GX(-1) + 1] holds.
Below are additional follow-up questions
Could the expression GX(-1) + 1 be zero or negative, and what would that imply?
If GX(-1) + 1 ≤ 0, then by the relationship P(X is even) = ½ [GX(-1) + 1], we would get a non-positive or even negative probability for “X is even.” A probability cannot be negative, so clearly this scenario cannot arise for any valid probability distribution with finite or at least absolutely summable probabilities.
In detail, recall that GX(-1) = Σ pn (-1)n. Each pn is nonnegative, and Σ pn = 1. For GX(-1) + 1 to be non-positive, you would need GX(-1) ≤ -1. For typical discrete distributions with exponentially decaying tails, GX(-1) will usually lie in the range (−1, 1). Even if parts of the sum are negative due to alternating signs, it is very unlikely for the total sum to be less than or equal to −1 unless there’s a pathological case of extremely large probabilities on odd indices. But even then, the probabilities sum to 1, limiting how negative the alternating sum can become.
In practice, for a legitimate PMF that sums to 1, GX(-1) + 1 must be nonnegative. Hence, P(X is even) remains valid and cannot exceed 1 or fall below 0.
What happens if the random variable X only takes even values?
If X takes only even values (for instance, X = 0, 2, 4, … with certain probabilities summing to 1), then P(X is even) = 1.
From the relationship P(X is even) = ½ [GX(-1) + 1], we get 1 = ½ [GX(-1) + 1]. This implies GX(-1) + 1 = 2, so GX(-1) = 1.
To see how that arises explicitly, consider that for any n that is not even, pn = 0. So the alternating sum GX(-1) = Σ p2n (−1)2n = Σ p2n = 1. Hence, the formula is consistent for the degenerate “all-even” case.
Could there be numerical instability issues when computing GX(-1) for large support?
Yes, particularly if the random variable X has a large range of values (either finite but large or infinite). The alternating sum GX(-1) = Σ pn (−1)n might cause significant numerical cancellation. For instance, if p2n and p2n+1 have similar magnitudes, the terms in the sum can subtract from each other, leading to precision loss when done in floating-point arithmetic.
In real-world computations, you might handle this by summing even and odd terms separately in double precision (e.g., accumulate Σ p2n and Σ p2n+1 independently, then subtract them), or by using arbitrary-precision libraries if you need extremely high precision. You also want to ensure that summation orders are chosen to reduce catastrophic cancellation.
How do we verify the relationship if the distribution pn is only known empirically?
Sometimes you have only sample data for X and do not have an explicit PMF. In that case, you would estimate pn by the empirical frequency of each n in your sample, say from N observations:
Let &hat;pn = count(n in sample) / N.
Then compute &hat;GX(-1) = Σ &hat;pn (−1)n.
Estimate the probability of even outcomes by summing the empirical frequencies of even n, i.e., Σ &hat;p2n.
In large samples, these two ways of estimating P(X is even) should match well: Σ &hat;p2n ≈ ½ [ &hat;GX(-1) + 1 ].
Discrepancies would reflect sampling variability. One would expect them to converge as N grows, by the law of large numbers.
Can we generalize this technique to compute P(X mod m = k) for other values of m?
Yes. The idea can be generalized using roots of unity. For instance, to find P(X mod 3 = 0), you could use the fact that substituting complex cube roots of unity into GX(z) can help separate the probabilities into residue classes mod 3. Concretely, let ω be a primitive 3rd root of unity. Then:
GX(ω0) = Σ pn (ω0)n = Σ pn, which is 1.
GX(ω) = Σ pn ωn.
GX(ω2) = Σ pn ω2n.
By combining these expressions appropriately, you can extract Σ p3n, Σ p3n+1, and Σ p3n+2. The case z = −1 is simply the 2nd root of unity scenario (since −1 is eiπ).
Hence, the specific formula for P(X is even) is a special case of a broader approach using roots of unity to isolate probabilities associated with certain residue classes.
If we had a shifted random variable Y = X + c, does the formula change?
If Y = X + c for some integer shift c, then the event “Y is even” becomes “X + c is even,” which is the same as “X is even if c is even, or X is odd if c is odd.”
If c is even, P(Y is even) = P(X is even).
If c is odd, P(Y is even) = P(X is odd) = 1 − P(X is even).
Meanwhile, the generating function for Y is zc GX(z) if c ≥ 0, because Y shifts the distribution by c. Evaluating that new generating function at z = −1 introduces an extra factor of (−1)c. That factor will flip the sign accordingly and yield the appropriate relationship for the evenness or oddness.
Is there a simple check to see if GX(z) is valid near z = −1 for a heavy-tailed distribution?
One way is to test absolute summability of pn. If Σ |pn (−1)n| = Σ pn < ∞, then GX(-1) is absolutely convergent. For simpler distributions like geometric, binomial, Poisson, negative binomial, and so on, you can be fairly confident GX(-1) converges because these distributions have exponential or factorial-type decays. For extremely heavy-tailed distributions (e.g., pn ~ n−α for α ≤ 1), the series might fail to converge absolutely. You could still have conditional or alternating series convergence, but you should check the conditions of the alternating series test or Dirichlet’s test.
If convergence fails, you cannot simply plug in z = −1 and apply the identity. You might need a different approach (perhaps partial sums or limiting processes) to define P(X is even).
How to extend the idea to a bivariate distribution?
If you have two nonnegative integer-valued random variables, say (X, Y), you can still define a joint generating function GX,Y(z, w) = Σ Σ P(X = m, Y = n) zm wn. Evaluating at (z, w) = (−1, −1) gives Σ Σ P(X = m, Y = n) (−1)m+n. From that, you could isolate probabilities of certain parity combinations, like P(X+Y is even), by grouping terms where m + n is even vs. odd.
In particular, m + n is even if both m and n are even or both are odd. Then:
(−1)m+n = +1 for m+n even,
(−1)m+n = −1 for m+n odd.
So GX,Y(−1, −1) = P(X+Y is even) − P(X+Y is odd).
From here, you can argue similarly that P(X+Y is even) = ½ [ GX,Y(−1, −1) + 1 ]. This is consistent with the univariate case but extended to two dimensions, showing how powerful the approach can be in a multi-dimensional setting.
Could we directly relate the parity result to a known distribution’s PMF without explicitly writing the entire PMF?
Yes, if there is a closed-form generating function. For instance, for X ~ Binomial(n, p), the PGF is GX(z) = (1 − p + pz)n. Evaluating at z = −1 gives GX(−1) = (1 − p − p)n = (1 − 2p)n. Then:
P(X is even) = ½ [(1 − 2p)n + 1].
This single compact expression spares you from computing p0, p1, …, pn individually. Similar shortcuts exist for Poisson, negative binomial, or other distributions once you know the closed-form generating function.
Notably, these shortcuts provide a very efficient way to handle parity questions in large-scale computations where you’d otherwise need to sum large, complicated PMFs.