Probability and Statistics — Distributions
🟢 Lite — Quick Review (1h–1d)
Rapid summary for last-minute revision before your exam.
Distributions to Memorise for GATE
| Distribution | PMF/PDF | Mean | Variance |
|---|---|---|---|
| Binomial | $P(X=r) = \binom nr p^r q^{n-r}$ | $np$ | $npq$ |
| Poisson | $P(X=r) = \dfrac{e^{-\lambda}\lambda^r}{r!}$ | $\lambda$ | $\lambda$ |
| Normal | $f(x) = \dfrac{1}{\sigma\sqrt{2\pi}}e^{-(x-\mu)^2/2\sigma^2}$ | $\mu$ | $\sigma^2$ |
| Exponential | $f(x)=\lambda e^{-\lambda x}, x\ge0$ | $\dfrac{1}{\lambda}$ | $\dfrac{1}{\lambda^2}$ |
Key formulas at a glance:
- Bayes’ Theorem: $P(A|B) = \dfrac{P(B|A)\cdot P(A)}{P(B)}$
- Conditional Probability: $P(A\cap B) = P(A)\cdot P(B|A)$
- Standard Normal: $Z = \dfrac{X-\mu}{\sigma}$, use $Z$-tables
- Poisson as limit of Binomial: when $n\to\infty$, $p\to0$, $np=\lambda$
⚡ GATE trap: Students often confuse “memoryless” property — it belongs to the exponential distribution, not the normal or Poisson. The exponential is the only continuous distribution that is memoryless.
⚡ GATE shortcut: For normal distribution problems, always standardise to $Z$ first. GATE questions usually give $P(Z<2)$ style values in the question itself or expect you to use symmetry $P(|Z|<a)=2P(Z<a)-1$.
🟡 Standard — Regular Study (2d–2mo)
Standard content for students with a few days to months.
Binomial Distribution
$X\sim B(n,p)$ models $n$ independent Bernoulli trials, each with success probability $p$.
$$P(X=r) = \binom{n}{r} p^r (1-p)^{n-r}, \quad r=0,1,\dots,n$$
- Mean: $\mu = np$ | Variance: $\sigma^2 = np(1-p)$
- Mode: $\lfloor (n+1)p \rfloor$ (most likely number of successes)
- Additive property: If $X_1 \sim B(n_1,p)$ and $X_2 \sim B(n_2,p)$ independently, then $X_1+X_2 \sim B(n_1+n_2,p)$
When to use it: Fixed number of trials, two outcomes per trial, constant $p$.
Poisson Distribution
$X\sim P(\lambda)$ is a limit of binomial as $n\to\infty$, $p\to0$, $np\to\lambda$.
$$P(X=r) = \frac{e^{-\lambda}\lambda^r}{r!}, \quad r=0,1,2,\dots$$
- Mean = Variance = $\lambda$ (this is unique among common distributions)
- Additive: If $X_1\sim P(\lambda_1)$, $X_2\sim P(\lambda_2)$ independent, then $X_1+X_2\sim P(\lambda_1+\lambda_2)$
- Used for rare events: arrivals per unit time, defects per unit area, etc.
GATE Question Pattern (2019, 2016): “If $X$ and $Y$ are independent Poisson with parameters 3 and 5, find $P(X+Y=4)$.” Use additive property → $P(X+Y=4) = \dfrac{e^{-8}\cdot 8^4}{4!}$.
Normal Distribution
$X\sim N(\mu,\sigma^2)$ is the most important continuous distribution in GATE.
$$f(x) = \frac{1}{\sigma\sqrt{2\pi}} \exp!\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)$$
- Bell-shaped, symmetric about $\mu$
- 68-95-99.7 rule: $P(\mu\pm\sigma)\approx68%$, $P(\mu\pm2\sigma)\approx95%$, $P(\mu\pm3\sigma)\approx99.7%$
- Standard normal $Z\sim N(0,1)$: Always standardise: $Z = \dfrac{X-\mu}{\sigma}$
- Linear combinations: Any linear combination of independent normal variables is also normal.
GATE shortcut: Use $Z$-table values — common ones to memorise:
- $P(Z<1)=0.8413$, $P(Z<1.645)=0.95$, $P(Z<2)=0.9772$, $P(Z<2.576)=0.995$
Exponential Distribution
$X\sim \text{Exp}(\lambda)$ models waiting time until the first event in a Poisson process.
$$f(x) = \lambda e^{-\lambda x}, \quad x\ge 0; \quad F(x)=1-e^{-\lambda x}$$
- Mean: $\frac{1}{\lambda}$ | Variance: $\frac{1}{\lambda^2}$
- Memoryless property: $P(X>s+t | X>s) = P(X>t)$
- Relationship: If $N(t)\sim P(\lambda t)$ counts events, the inter-arrival time $\sim \text{Exp}(\lambda)$
Bayes’ Theorem and Conditional Probability
$$P(A|B) = \frac{P(B|A)\cdot P(A)}{P(B)}, \quad P(B) = \sum_i P(B|A_i)P(A_i)$$
GATE frequently combines Bayes with discrete distributions. Example pattern: “A manufacturing process produces $1%$ defective items. A test detects a defective with probability $0.99$ and a non-defective with probability $0.95$. If an item tests positive, find the probability it is actually defective.” Use Bayes on events $D$ (defective) and $T$ (tests positive).
Common GATE Mistakes to Avoid
- Confusing binomial ($n$ finite, $p$ constant) with Poisson (rare events limit)
- Forgetting to standardise normal before using $Z$-tables
- Applying memoryless property to wrong distribution
- Using variance formula for binomial as $p(1-p)$ instead of $np(1-p)$
- Mixing up $P(A|B)$ and $P(B|A)$ in Bayes’ calculations
🔴 Extended — Deep Study (3mo+)
Comprehensive coverage for students on a longer study timeline.
Binomial Distribution — Deep Dive
The binomial distribution arises from the sum of $n$ independent Bernoulli$(p)$ trials:
$$X = \sum_{i=1}^n X_i, \quad X_i \in {0,1}, \quad P(X_i=1)=p$$
Moment generating function (MGF): $$M_X(t) = (pe^t + q)^n, \quad q=1-p$$
Recurrence relation for probabilities: $$\frac{P(X=r)}{P(X=r-1)} = \frac{(n-r+1)p}{rq}, \quad r\ge 1$$
This recurrence lets you compute $P(X=r)$ iteratively without factorials — useful when $n$ is large and $p$ is moderate.
Mode derivation: The ratio $P(X=r)/P(X=r-1) > 1$ when $r < (n+1)p$. So mode is either $\lfloor(n+1)p\rfloor$ or $(n+1)p-1$ (when $(n+1)p$ is integer, both $r$ and $r-1$ are modes).
Approximation relationships:
- Poisson approximation: When $n\ge 20$ and $p\le 0.05$, $B(n,p)\approx P(np)$ works well.
- Normal approximation: When $np>5$ and $n(1-p)>5$, $B(n,p)\approx N(np, np(1-p))$ with continuity correction.
Continuity correction: $P(a\le X\le b) \approx P!\left(\frac{a-0.5-np}{\sqrt{npq}} \le Z \le \frac{b+0.5-np}{\sqrt{npq}}\right)$
Poisson Distribution — Deep Dive
The Poisson process satisfies three axioms: (1) independent increments, (2) stationary increments, (3) $P(N(h)\ge 2)=o(h)$. Under these, $N(t)\sim P(\lambda t)$.
Sums of independent Poissons are Poisson — this is frequently used in GATE problems involving combined arrival rates.
Conditional distribution: If $X_1\sim P(\lambda_1)$ and $X_2\sim P(\lambda_2)$ are independent, then: $$X_1 | (X_1+X_2=n) \sim \text{Binomial}!\left(n, \frac{\lambda_1}{\lambda_1+\lambda_2}\right)$$
This is a negative binomial result as well — it confirms the Poisson-Dirichlet structure.
GATE 2020 style question: “If $X$ and $Y$ are independent Poisson random variables with means 2 and 4 respectively, find $E[(X+Y)^2]$.” Solution: $E[(X+Y)^2] = \text{Var}(X+Y) + [E(X+Y)]^2 = (2+4) + (6)^2 = 6+36 = 42$.
Normal Distribution — Deep Dive
The normal distribution’s MGF: $M_X(t) = \exp(\mu t + \frac{1}{2}\sigma^2 t^2)$
Linear transformations preserve normality:
- If $X\sim N(\mu,\sigma^2)$, then $aX+b \sim N(a\mu+b, a^2\sigma^2)$
- Standardisation: $Z = \dfrac{X-\mu}{\sigma} \sim N(0,1)$
Central Limit Theorem connection: For any distribution with mean $\mu$ and variance $\sigma^2$, $\dfrac{\bar{X}-\mu}{\sigma/\sqrt{n}} \to N(0,1)$ as $n\to\infty$. This is why normal appears everywhere in statistics.
Skewness and kurtosis:
- Normal has skewness = 0, kurtosis excess = 0
- Any departure from these signals non-normality
Chi-square as a square of normals: If $Z_1,\dots,Z_k$ are iid $N(0,1)$, then $\chi^2_k = Z_1^2+\cdots+Z_k^2 \sim \text{Chi-square}(k)$. This links the normal to the chi-square distribution.
Exponential Distribution — Deep Dive
Derivation from Poisson: Inter-arrival times in a Poisson$(\lambda)$ process are iid $\text{Exp}(\lambda)$.
Memoryless proof: $$P(X>s+t|X>s) = \frac{P(X>s+t)}{P(X>s)} = \frac{e^{-\lambda(s+t)}}{e^{-\lambda s}} = e^{-\lambda t} = P(X>t)$$
Rate vs mean: Be careful — the parameter $\lambda$ in $f(x)=\lambda e^{-\lambda x}$ is the rate, so mean $=1/\lambda$. Some textbooks use the scale parameter $\beta=1/\lambda$ instead, writing $f(x)=\frac{1}{\beta}e^{-x/\beta}$.
Minimum of exponentials: If $X_1,\dots,X_n$ are independent $\text{Exp}(\lambda_i)$, then $\min(X_1,\dots,X_n)\sim \text{Exp}(\sum \lambda_i)$.
Bayes’ Theorem — Applications
Continuous version: If $\theta$ has a prior density $\pi(\theta)$ and data $x$ has likelihood $f(x|\theta)$, then: $$\pi(\theta|x) = \frac{f(x|\theta)\pi(\theta)}{\int f(x|\theta)\pi(\theta),d\theta}$$
This is the foundation of Bayesian inference, though GATE focuses on the discrete/finite case.
GATE 2018 hint: Bayes + Binomial appears as “posterior probability” problems. Practice: $P(\theta=0.3|X=5)$ where $X\sim B(10,\theta)$ and $\theta$ has prior $P(\theta=0.3)=P(\theta=0.5)=0.5$.
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