Probability That Uniform Plus Exponential Exceeds One
Let $U \sim \text{Uniform}(0, 1)$ and $V \sim \text{Exponential}(1)$ be independent random variables. Find $P(U + V > 1)$ to the nearest thousandth.
Hints
- With two non-identically-distributed random variables, condition on the simpler one. Here, conditioning on $U = u$ reduces the problem to a single tail probability for the exponential.
- For $V \sim \text{Exponential}(1)$, the tail probability is $P(V > x) = e^{-x}$ for $x \ge 0$. After conditioning, $P(V > 1 - u)$ is immediate.
- Apply the law of total probability: $P(U + V > 1) = \int_0^1 e^{-(1-u)} \, du$. Factor out $e^{-1}$ and integrate $e^u$ over $[0, 1]$.
Worked Solution
How to Think About It: $U$ is bounded on $[0, 1]$ while $V$ is unbounded on $[0, \infty)$. Since $V$ has no upper bound, conditioning on $U$ is the cleaner move -- once $U = u$ is fixed, we need $V > 1 - u$, and tail probabilities of the exponential are trivial to compute. The law of total probability handles the conditioning cleanly.
Quick Estimate: The expected value of $U + V$ is $E[U] + E[V] = 0.5 + 1 = 1.5$. The sum exceeds its own mean about half the time for symmetric distributions, but $U + V$ is skewed right (exponential has a long right tail). So we expect the probability to be noticeably above $0.5$. A rough guess: around $0.6$-$0.7$.
Approach: Condition on $U$ and integrate using the exponential tail probability.
Formal Solution:
By the law of total probability, conditioning on $U = u$: $$P(U + V > 1) = \int_0^1 P(V > 1 - u) \cdot f_U(u) \, du$$
Since $U \sim \text{Uniform}(0,1)$, the density is $f_U(u) = 1$ on $[0, 1]$.
For $V \sim \text{Exponential}(1)$, the survival function is $P(V > x) = e^{-x}$ for $x > 0$.
Since $u \in [0, 1]$, we have $1 - u \in [0, 1] \subset [0, \infty)$, so: $$P(V > 1 - u \mid U = u) = e^{-(1-u)}$$
Substituting: $$P(U + V > 1) = \int_0^1 e^{-(1-u)} \, du$$
Let $w = 1 - u$, $dw = -du$; limits go from $w = 1$ to $w = 0$: $$= \int_1^0 e^{-w} (-dw) = \int_0^1 e^{-w} \, dw = \left[-e^{-w}\right]_0^1 = -e^{-1} + e^0 = 1 - e^{-1}$$
Alternatively, directly: $$\int_0^1 e^{-(1-u)} \, du = e^{-1}\int_0^1 e^{u} \, du = e^{-1}[e^u]_0^1 = e^{-1}(e - 1) = 1 - e^{-1}$$
Numerically: $$P(U + V > 1) = 1 - e^{-1} \approx 1 - 0.3679 = \boxed{0.632}$$
Answer: $P(U + V > 1) = 1 - e^{-1} \approx 0.632$.
Intuition
Conditioning on the bounded variable is a standard trick when one random variable is bounded and the other is not. It converts a joint probability over two variables into a simple integral of a one-dimensional tail probability -- which for the exponential is just $e^{-x}$, making the integral trivial.
The answer $1 - e^{-1} \approx 0.632$ is the same as $P(V > 1)$ for a single Exponential(1) variable -- not a coincidence. Geometrically, $P(U + V > 1)$ is the integral of the exponential density over the region $\{(u, v) : u + v > 1, u \in [0,1], v > 0\}$, which can be computed as the complement of the region below the line $u + v = 1$. This type of calculation -- integrating over a half-plane in a two-dimensional density -- comes up in options pricing (probability of joint events) and in risk calculations (probability that a portfolio loss exceeds a threshold given factor realizations).