←Back to the bookChapter 55 · Hardware Lifespans (Exponential)

Chapter 55 · Case Study

Hardware Lifespans: the Exponential ⏳

How long until a component fails? Time between random failures follows the exponential distribution. We fit the failure rate, read reliability off the CDF, and confront its strange memoryless property.

Statistics, Data Science and AI: A Visual Handbook · John Fisher · 2026

⚙️ Setup & data

import numpy as np, pandas as pd, matplotlib.pyplot as plt
from scipy import stats
AMBER="#d97706"; TEAL="#0d9488"; INK="#1a2138"; GRID="#e6e9f2"; PINK="#db2777"
plt.rcParams.update({"figure.facecolor":"white","axes.facecolor":"white","figure.dpi":110,"font.size":11,
   "axes.edgecolor":GRID,"axes.grid":True,"grid.color":GRID,"axes.axisbelow":True,"axes.spines.top":False,
   "axes.spines.right":False,"axes.titlesize":12,"axes.titleweight":"bold","legend.frameon":False})
BASE="https://raw.githubusercontent.com/johnfisher-ai/Statistics-Data-Science-AI-Visual-Book/main/data/"
try:    df = pd.read_csv("../../data/hardware_component_lifespans.csv")
except FileNotFoundError: df = pd.read_csv(BASE+"hardware_component_lifespans.csv")
print("loaded:", df.shape)
df.head()

loaded: (1000, 4)

	component_serial	manufacture_batch	component_type	lifespan_years
0	SN_880000	BATCH_109	NVMe_Controller	11.245
1	SN_880001	BATCH_113	NVMe_Controller	1.154
2	SN_880002	BATCH_115	SSD_Drive	2.691
3	SN_880003	BATCH_136	NVMe_Controller	1.867
4	SN_880004	BATCH_110	NVMe_Controller	7.741

BEAT 1 · FIT THE FAILURE RATE

One rate, lambda

The exponential models the waiting time between Poisson events, here, between failures. Its rate lambda is one over the mean lifespan, and its right-skewed PDF lambda e^(-lambda x) drops onto the data.

x = df["lifespan_years"]
rate = 1/x.mean()
print(f"components : {len(x):,}")
print(f"mean lifespan = {x.mean():.3f} years")
print(f"failure rate lambda = 1/mean = {rate:.4f} per year")
print(f"exponential mean = 1/lambda = {1/rate:.3f}")
print(f"exponential std  = 1/lambda = {1/rate:.3f}  (mean = std, right-skewed)")
print(f"median = ln(2)/lambda = {np.log(2)/rate:.3f} years  (< mean, the skew)")

components : 1,000
mean lifespan = 4.972 years
failure rate lambda = 1/mean = 0.2011 per year
exponential mean = 1/lambda = 4.972
exponential std  = 1/lambda = 4.972  (mean = std, right-skewed)
median = ln(2)/lambda = 3.446 years  (< mean, the skew)

fig,ax=plt.subplots(figsize=(7,3.6))
ax.hist(x,bins=40,density=True,color=AMBER,alpha=0.7,label="observed lifespans")
xs=np.linspace(0,x.max(),300); ax.plot(xs,stats.expon.pdf(xs,scale=1/rate),color=TEAL,lw=2.5,label=f"Exponential(lambda={rate:.2f})")
ax.axvline(x.mean(),color=PINK,ls="--",lw=2,label=f"mean = {x.mean():.1f} yr")
ax.set_xlabel("lifespan (years)"); ax.set_ylabel("density"); ax.set_title("Component lifespans: observed vs exponential"); ax.legend()
plt.tight_layout(); plt.show()

# Formal goodness-of-fit: does the Exponential really fit? KS test + an exponential QQ plot.
ks = stats.kstest(x, "expon", args=(0, 1/rate))
print(f"Kolmogorov-Smirnov vs Exponential: D = {ks.statistic:.3f}, p = {ks.pvalue:.3f}  -> "
      + ("cannot reject the fit (Exponential is a good model)" if ks.pvalue>0.05 else "reject: not exponential"))
fig, ax = plt.subplots(figsize=(4.8,3.8))
stats.probplot(x, dist=stats.expon, sparams=(0, 1/rate), plot=ax)
ax.get_lines()[0].set_color(AMBER); ax.get_lines()[0].set_markersize(3)
ax.set_title("Exponential QQ plot (points on the line = good fit)"); plt.tight_layout(); plt.show()

Kolmogorov-Smirnov vs Exponential: D = 0.017, p = 0.935  -> cannot reject the fit (Exponential is a good model)

The overlay looks right, but a curve on a histogram is easy to talk yourself into. Two honest checks earn the model: the QQ plot lines the data's quantiles up against an exponential's (points on the diagonal = a good fit, and here they sit on it), and the formal Kolmogorov-Smirnov test puts a number on the gap between the data and the fitted exponential (D = 0.02, p = 0.94, so we cannot reject it). Now the Exponential is validated, not just assumed.

The exponential's right-skew fits: most parts fail early-ish, a few last a long time. Mean and standard deviation are both 5 years (their equality is a signature), and the median, ln(2)/λ ≈ 3.4 years, sits below the mean, the mark of a skewed distribution.

BEAT 2 · RELIABILITY FROM THE CDF

Early-fatigue risk

Reliability lives in the CDF: P(fails before x years) = 1 - e^(-lambda x), and its complement, the survival function e^(-lambda x), is the chance a part is still alive at age x.

for yr in [1,2,5,10]:
    print(f"P(fails before {yr:2d} yr) = 1 - e^(-lambda*{yr}) = {stats.expon.cdf(yr,scale=1/rate):.4f}   survival {stats.expon.sf(yr,scale=1/rate):.4f}")
print(f"\nobserved early-fatigue P(<1 yr) = {(x<1).mean():.4f}")
print(f"warranty length for 5% early-failure = {stats.expon.ppf(0.05,scale=1/rate):.3f} years")

P(fails before  1 yr) = 1 - e^(-lambda*1) = 0.1822   survival 0.8178
P(fails before  2 yr) = 1 - e^(-lambda*2) = 0.3312   survival 0.6688
P(fails before  5 yr) = 1 - e^(-lambda*5) = 0.6342   survival 0.3658
P(fails before 10 yr) = 1 - e^(-lambda*10) = 0.8662   survival 0.1338

observed early-fatigue P(<1 yr) = 0.1800
warranty length for 5% early-failure = 0.255 years

About 18% of components fail within the first year, the early-fatigue risk, matched by the data. The survival function says 37% are still running at 5 years (= e⁻¹, since 5 years is one mean lifetime). To cap early failures at 5%, the warranty should run only about 0.26 years, the CDF turns a risk target into a policy.

BEAT 3 · MEMORYLESS: NO WEAR-OUT

A 5-year-old part is as good as new

The exponential is memoryless: P(survives t more | survived s) = P(survives t). Its hazard rate is CONSTANT, parts do not age. We confirm by simulation, and note where the model breaks.

rng = np.random.default_rng(55)
sim = rng.exponential(1/rate, size=2_000_000)
base = stats.expon.sf(1,scale=1/rate)
cond = np.mean(sim[sim>5] > 6)        # survived 5 years: chance of 1 more?
print(f"P(survive 1 more year, fresh part)      = {base:.4f}")
print(f"P(survive 1 more year | already 5 yr)   = {cond:.4f}  (same -> memoryless)")
print("Constant hazard: the exponential assumes no wear-out. Real parts that DO wear out need a Weibull.")

P(survive 1 more year, fresh part)      = 0.8178
P(survive 1 more year | already 5 yr)   = 0.8177  (same -> memoryless)
Constant hazard: the exponential assumes no wear-out. Real parts that DO wear out need a Weibull.

Whether a component is brand new or has run for five years, its chance of lasting one more year is the same 82%: the exponential has no memory and a constant hazard rate. That is its great simplification, and its limitation: parts that genuinely wear out (rising hazard) call for the Weibull distribution. Knowing when memorylessness is and is not realistic is the engineer's judgment call.

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✅ Case closed

Time-to-failure is exponential. We fit the rate λ ≈ 0.20/year (mean 5 years), read early-fatigue risk and survival off the CDF, set a warranty from a risk target, and confronted the memoryless, constant-hazard assumption. The exponential is the continuous twin of the geometric, the waiting time of the Poisson.

Statistics, Data Science and AI: A Visual Handbook · © 2026 John Fisher

←Back to the bookChapter 55 · Hardware Lifespans (Exponential)