ML Interview Q Series: Calculating E(X|X>Y) and E(Y|X>Y) for Independent Exponential Random Variables.

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Let (X) and (Y) be two independent random variables that have the same exponential density function with expected value (1/\lambda). What are (\displaystyle E(X \mid X > Y)) and (\displaystyle E(Y \mid X > Y))?

Short Compact solution

Using the fact that (X) and (Y) are i.i.d. exponential with parameter (\lambda), it follows that

Comprehensive Explanation

Setup and Key Distributions

Suppose (X) and (Y) are i.i.d. with exponential density function fX(x) = lambda * exp(-lambda x) for x>0 and similarly for Y.

The exponential distribution with rate lambda has expected value 1/lambda. Because the variables are i.i.d., we can leverage symmetry properties. We also know that for i.i.d. exponentials X and Y, the event X>Y occurs with probability 1/2.

Deriving E[X | X > Y]

1. Distribution of X given X>Y:

   We can compute the conditional density of X given X>Y by recognizing that: P(X ≤ x, X>Y) = ∫[0 to x] fX(u) (∫[0 to u] fY(y) dy) du and then dividing by P(X>Y) (which is 1/2). A more direct argument uses: conditional pdf of X given X>Y = 2 * fX(x) * ∫[0 to x] fY(y) dy

   Substituting fX(x)=lambda e^(-lambda x) and fY(y)=lambda e^(-lambda y), we get ∫[0 to x] fY(y) dy = ∫[0 to x] lambda e^(-lambda y) dy = 1 - e^(-lambda x). Hence the conditional pdf of X (for x>0) given X>Y is 2 * lambda e^(-lambda x) * [1 - e^(-lambda x)].

2. Expected value calculation:

   We compute E[X | X>Y] by integrating x times the above conditional density: E[X | X>Y] = ∫[0 to ∞] x * [2 * lambda e^(-lambda x) (1 - e^(-lambda x)) ] dx.

   A more straightforward approach (without doing this integral explicitly by parts) uses the memoryless property of the exponential distribution. However, carrying out the integration shows that the result simplifies to 3/(2 lambda).

Deriving E[Y | X>Y]

1. Distribution of Y given X>Y:

   By symmetry, or by a similar integral approach, the conditional pdf of Y given X>Y is 2 * fY(y) * ∫[y to ∞] fX(x) dx for y>0. Because ∫[y to ∞] fX(x) dx = e^(-lambda y), this becomes 2 * lambda e^(-lambda y) * e^(-lambda y) = 2 * lambda e^(-2 lambda y), although one can confirm exact functional forms by direct substitution.

2. Expected value calculation:

   E[Y | X>Y] = ∫[0 to ∞] y * [2 * lambda e^(-lambda y) ∫[y to ∞] fX(x) dx ] dy, which evaluates to 1/(2 lambda). Again, symmetry arguments and the memoryless property can be invoked to verify this more directly.

Hence, the final results:

These quantities are consistent with the well-known property that when two i.i.d. exponential random variables are compared, the larger one on average is 3/(2 lambda) while the smaller one is 1/(2 lambda).

Follow-up question: Why is P(X>Y) = 1/2 for i.i.d. exponential variables?

Because X and Y have the same continuous distribution and are independent, the probability that X>Y must be 1/2 by symmetry. In general, for two i.i.d. continuous random variables, P(X>Y)=1/2 (and P(X=Y)=0 for a continuous distribution).

Follow-up question: How does the memoryless property help in deriving these expectations?

The exponential distribution is memoryless, meaning that for x>0 and t>0, P(X> x + t | X> x) = P(X> t). Intuitively, once you know X is already bigger than Y, you can think of the tail beyond Y as a “fresh start.” Many integrals that might otherwise be complicated can be broken down using this memoryless property. For instance, after establishing that X>Y, we can express X=Y+(X−Y) and use the property that X−Y is still exponentially distributed (or at least analyze how that distribution shifts).

Follow-up question: What if X and Y were not exponentially distributed?

If X and Y were i.i.d. with some other distribution, the memoryless property would not hold, so we cannot simply assert simple expressions for E[X|X>Y]. The result would require computing integrals of the conditional density, and it would not in general produce as elegant a ratio between E(X|X>Y) and E(X). The computation would still be conceptually straightforward (using the definition of conditional expectation and pdfs), but lacking a memoryless property typically means the integrals are more involved and do not reduce to simple constants times 1/lambda.

Follow-up question: How might one verify these results numerically in Python?

A quick empirical simulation can demonstrate the correctness of 3/(2 lambda) and 1/(2 lambda). One could write:

import numpy as np

lambda_val = 2.0  # example rate
N = 10_000_000

X = np.random.exponential(1/lambda_val, N)
Y = np.random.exponential(1/lambda_val, N)

mask = X > Y  # boolean mask where X>Y

print("Empirical E[X | X>Y]:", X[mask].mean())
print("Empirical E[Y | X>Y]:", Y[mask].mean())

Running this code would yield approximate values close to 3/(2 lambda_val) for X and 1/(2 lambda_val) for Y, confirming the analytical result.

Below are additional follow-up questions

How might these results change if we only observe a truncated version of X and Y?

If X and Y were observed only when they exceed a certain threshold m (for instance, due to sensor limitations), we would effectively be dealing with truncated exponentials. Then, the conditional distributions of X and Y given X>m and Y>m would be shifted and scaled compared to the standard exponential. Specifically, each variable would have a truncated density:

fX_truncated(x) = (lambda e^(-lambda x)) / (e^(-lambda m)) for x>m

A key subtlety arises in the event X>Y. Because we only observe values above m, we would need to recast X>Y in that truncated domain. We cannot simply use the standard property P(X>Y)=1/2, as X and Y are restricted to x>m, y>m. The memoryless property is still valid for exponentials, but with the condition that X>m and Y>m, the event X>Y is altered by the truncation. The expectation E(X | X>Y) under this truncated scenario would involve:

1. Conditioning on X>m, Y>m.

2. Applying the memoryless property for x>m and y>m to identify how the distribution shifts.

3. Adjusting integrals to account for the domain m to infinity.

In practice, ignoring the truncation when computing these conditional expectations could lead to biased estimates of E(X | X>Y) and E(Y | X>Y). A real-world pitfall might occur if sensor data cannot record values below m, yet the standard formula 3/(2 lambda) is mistakenly used.

Is there a convenient way to handle correlations if X and Y are not independent?

When X and Y are correlated, we can no longer rely on the factorization fX(x)*fY(y). Instead, the joint distribution fX,Y(x,y) must be used. The event X>Y then has probability

P(X>Y) = ∫∫_{ x>y } fX,Y(x,y) dx dy

and the conditional density would become:

fX|X>Y(x | X>Y) = fX,Y(x,y) / P(X>Y), integrated appropriately over y<x.

This can become considerably more complex, especially if the correlation structure is intricate (e.g., a Gaussian copula or other forms of dependence). A potential pitfall is to assume that the probability X>Y remains at 1/2 or that the conditional expectations preserve the same symmetry found in the independent case. One must handle the correlation carefully in any real-world modeling scenarios, possibly by writing down and integrating over the joint distribution or by simulation-based approaches (e.g., Monte Carlo sampling).

How does the result extend to the maximum and minimum of multiple exponential variables?

When there are more than two i.i.d. exponential random variables, say X1, X2, ..., Xn, and we look at the event that Xk is the maximum (or minimum) among all Xi, we can use the property that the minimum of i.i.d. exponentials is also exponential with rate n * lambda, and the index of the minimum is uniformly distributed among them. For the maximum, we typically rely on the complementary events or the fact that the distribution of the maximum among n exponentials is related to the sum of n independent exponentials in a certain sense. However, the conditional expectation E(Xk | Xk is the maximum) does not simply become 3/(2 lambda) for n>2. Instead, one has to compute:

E(Xk | Xk > max(all other Xi)) = ∫ x fXk|Xk=max(x) dx

and the result is more involved, though it can still be worked out by using order statistics and exponential properties. In real-world scenarios, a pitfall is to treat the pairwise result 3/(2 lambda) as though it generalizes directly to n variables without carefully going through the order-statistics derivation.

Could these findings lead to biased estimates if we inadvertently use them for a mixture of exponentials?

If the distribution of X or Y is not pure exponential but rather a mixture, such as a mixture of exponentials with different rates (or a Gamma distribution that arises, for example, from summing multiple exponentials), then the memoryless property does not hold in general. If one incorrectly treats that mixture as a single exponential, the standard results E(X | X>Y)=3/(2 lambda) and E(Y | X>Y)=1/(2 lambda) would no longer be correct. A mixture can have a long tail in which large values are more likely than in a single exponential with a single rate. In practice, this discrepancy might manifest as repeated over- or underestimation of X or Y. It’s crucial to do a goodness-of-fit check (like a Q-Q plot or likelihood ratio test) before assuming a single exponential model.

What if the observed data is actually discrete, but one models it as exponential?

Sometimes data are effectively discrete (e.g., integer counts or time in discrete increments) but is approximated by a continuous distribution such as the exponential. This can introduce modeling errors. For discrete analogs (like the geometric distribution, which is memoryless for discrete counts), one might get analogous formulas, but they will not match the 3/(2 lambda) exactly. For instance, with geometric random variables that are i.i.d., the probability X>Y would still be 1/2 by symmetry, but the expectation E(X | X>Y) would not have as direct a closed-form. A pitfall is to rely on continuous approximation for discrete data without verifying that the time increments are sufficiently small to make the approximation valid.

How might we incorporate Bayesian updating into the scenario?

In a Bayesian perspective, if lambda is not known precisely but instead has a prior distribution, then after observing X>Y, the posterior for lambda might be updated. The event X>Y could shift our belief about the parameter. Specifically, if one uses a conjugate prior for lambda (a Gamma distribution), the observation that X>Y modifies the posterior according to Bayes’ rule:

posterior(lambda) ∝ prior(lambda) * P(X>Y | lambda)

However, we do not observe the exact values of X and Y, only that X>Y. That partial information does update lambda, though in a weaker way than full numeric values of (X, Y) would. A subtlety is that the memoryless property helps with the direct calculation of P(X>Y | lambda), but the real pitfall is ignoring partial data (the magnitude of X or Y) and only using the binary event X>Y, which can lead to a less informative posterior. The final estimates for E(X | X>Y) might shift if we incorporate the prior and observe many outcomes of X>Y or X<Y.

How does censoring affect the analysis?

Censoring occurs when we only know that a random variable exceeds a certain threshold but do not see its actual value. Suppose we only know that X>c for some c>0, and also X>Y is observed. Then the conditional probability and the associated conditional expectation revolve around events:

X>c and X>Y

One must carefully compute:

P(X>Y, X>c)

P(X>c)

and the corresponding conditional expectation. This can lead to complicated integrals because we have an extra condition X>c in addition to X>Y. Ignoring that second piece of information or conflating X>c with X>Y can create biased estimates in, for example, survival analysis or reliability modeling where incomplete data are common.

What happens if lambda is time-varying?

In some real-world processes (e.g., systems subject to fatigue or intermittent stress), the “rate” of occurrence for an exponential-like event might shift over time. This implies that X and Y might not truly be i.i.d. exponentials with a fixed parameter lambda but instead have a hazard rate function h(t) that changes. The memoryless property no longer applies if the hazard rate is not constant. So the result E(X | X>Y)=3/(2 lambda) breaks down for time-varying processes. One would need to consider the integrated hazard function and carefully adapt the approach:

1. Define an appropriate joint model for X(t) and Y(t).

2. Condition on X>Y by integrating the survival functions that reflect the time-varying hazard.

In practice, ignoring this time dependence is a major pitfall in domains like reliability engineering or medical event analysis, leading to misestimates of mean times to event or failure.

How do these formulas help in hypothesis testing about the rate lambda?

One might try to form a hypothesis test H0: lambda= lambda0 versus H1: lambda ≠ lambda0 based on a sample of pairs (X_i, Y_i). Under the null, we expect that whenever X_i>Y_i, the difference X_i - Y_i should follow an exponential distribution with the same rate lambda0 (by memorylessness). We can check whether the average of (X_i - Y_i) for all pairs in which X_i>Y_i matches 1/lambda0, or whether E(X | X>Y) aligns well with 3/(2 lambda0). A pitfall is that if data are not large enough, or if X and Y do not actually come from the same distribution, these test statistics might incorrectly reject or fail to reject the null. Additionally, if there are outliers or a mixture distribution, the test might be unreliable without robust modifications.

How do we handle real-world measurement errors in X and Y?

In practice, X and Y may be measured with random noise or measurement instruments that have calibration biases. Thus, even if the underlying phenomenon is i.i.d. exponential, the observed random variables might not strictly follow the same distribution. For example, if Y’s sensor under-reports by a constant offset or has a wider variance in measurement noise, it will influence the distribution of the observed Y-values and potentially alter the probability that (X>Y) in the measured data. This leads to biases in E(X | X>Y) and E(Y | X>Y) if we ignore measurement error. A robust approach might involve modeling measurement errors explicitly, specifying measurement error distributions, and then performing a deconvolution or measurement-error-corrected approach to estimate the true underlying X and Y. If measurement error is unaccounted for, one might incorrectly interpret the difference between observed X and Y as a real difference in the underlying signals.