Expected value, variance, standard deviation — properties: Exercises

4.9.4 · D4 · Maths › Probability Theory & Statistics › Expected value, variance, standard deviation — properties

Ek random variable $X$ (chance experiment se nikla ek number) ke do summaries neeche sab kuch mein kaam aate hain:

expected value $E [X]$ — probability mass ka balance point;
variance $Var (X) = E [X^{2}] - (E [X])^{2}$ — us balance point se average squared distance; aur iska square root standard deviation $σ = Var (X)$ .

Poora toolbox, ek card par taaki tumhe kabhi scroll na karna pade:

Level 1 — Recognition

Goal: spot karo ki kaunsa rule fire karta hai. Koi bhari algebra nahi.

Recall Solution 1.1

(a) Expectation linear hai, isliye constants seedhe nikal aate hain: $E [3 X - 2] = 3 E [X] - 2 = 3 (4) - 2 = 10.$ (b) Variance ke liye shift $- 2$ gayab ho jaata hai (poori picture ko left/right slide karne se spread nahi badlta) aur multiplier $3$ squared ho jaata hai: $Var (3 X - 2) = 3^{2} Var (X) = 9 \cdot 9 = 81.$ (c) Standard deviation square root hoti hai: $SD (3 X - 2) = 81 = 9$ . Equivalently $∣3∣ \cdot SD (X) = 3 \cdot 9 = 3 \cdot 3 = 9$ . ✓

Recall Solution 1.2

Ek 0/1 (indicator) variable ka $E [X] = p$ aur $Var (X) = p (1 - p)$ hota hai (dekho Bernoulli and Binomial Distributions). $E [X] = 0.2, Var (X) = 0.2 \cdot 0.8 = 0.16.$ $p (1 - p)$ kyun? Kyunki 0/1 variable ke liye $X^{2} = X$ hota hai, isliye $E [X^{2}] = p$ , aur $Var = E [X^{2}] - (E [X])^{2} = p - p^{2}$ .

Recall Solution 1.3

(a) False. Expectation ki linearity hamesha hold karti hai — dependent variables ke liye bhi. Ise kabhi independence ki zaroorat nahi padti. (b) True (essentially). Sum rule ek extra term $2 Cov (X, Y)$ carry karta hai. Independence $Cov = 0$ force karta hai, isliye do variances simply add ho jaate hain. (Strictly, uncorrelated hona kaafi hai — lekin independence ise guarantee karta hai.)

Level 2 — Application

Goal: real distributions par machine chalao.

Recall Solution 2.1

$E [X] = 1 (0.7) + 0 (0.3) = 0.7.$ $E [X^{2}] = 1^{2} (0.7) + 0^{2} (0.3) = 0.7 (kyunki X^{2} = X 0/1 variable ke liye) .$ $Var (X) = E [X^{2}] - (E [X])^{2} = 0.7 - 0.49 = 0.21.$ $σ = 0.21 \approx 0.4583.$

Recall Solution 2.2

Center of mass (har value apni probability se weighted): $E [X] = 0 (0.1) + 1 (0.2) + 2 (0.3) + 3 (0.4) = 0 + 0.2 + 0.6 + 1.2 = 2.0.$ Second moment LOTUS se $g (x) = x^{2}$ ke saath: $E [X^{2}] = 0 (0.1) + 1 (0.2) + 4 (0.3) + 9 (0.4) = 0 + 0.2 + 1.2 + 3.6 = 5.0.$ Variance computational formula se: $Var (X) = 5.0 - 2. 0^{2} = 5.0 - 4.0 = 1.0.$ $σ = 1.0 = 1.0.$

Recall Solution 2.3

Mean: $E [Y] = - 4 E [X] + 7 = - 4 (2.0) + 7 = - 1.$ Variance: multiplier square ho jaata hai (isliye sign gayab ho jaata hai) aur $+ 7$ shift kuch nahi karta: $Var (Y) = (- 4)^{2} Var (X) = 16 \cdot 1.0 = 16.$ SD: $σ_{Y} = ∣ - 4∣ \cdot σ_{X} = 4 \cdot 1.0 = 4.0.$ Absolute value wahi cheez hai jo $σ_{Y} \geq 0$ rakhti hai chahe $a = - 4$ negative ho.

Neeche ki figure dikhati hai kyun sign drop ho jaata hai aur shift spread ke liye invisible hai.

Figure — Expected value, variance, standard deviation — properties

Level 3 — Analysis

Goal: reason karo ki rule kab bend karta hai, aur combined variables ke baare mein.

Recall Solution 3.1

$X + X = 2 X$ , isliye scale rule se $Var (X + X) = Var (2 X) = 2^{2} Var (X) = 4 \cdot 5 = 20.$ Naively variances add karne se milta hai $Var (X) + Var (X) = 5 + 5 = 10$ . Ye isliye differ karte hain kyunki sum rule ko independence chahiye, aur $X$ apne aap se perfectly correlated hai. Missing piece hai $2 Cov (X, X) = 2 Var (X) = 2 \cdot 5 = 10$ , aur indeed $10 + 10 = 20$ . ✓

Recall Solution 3.2

$Var (a X + bY) = a^{2} Var (X) + b^{2} Var (Y) + 2 ab Cov (X, Y)$ use karo. (a) $a = b = 1$ : $Var (X + Y) = 4 + 9 + 2 (1) (1) (- 3) = 13 - 6 = 7.$ (b) $a = 1, b = - 1$ : cross term sign flip karta hai kyunki $2 ab = 2 (1) (- 1) = - 2$ : $Var (X - Y) = 4 + 9 + 2 (1) (- 1) (- 3) = 13 + 6 = 19.$ (c) Agar independent hote, toh $Cov = 0$ , isliye dono $4 + 9 = 13$ ho jaate. Dhyan do ki subtraction phir bhi variances add karta hai — spread accumulate hoti hai chahe tum random quantities add karo ya subtract.

Recall Solution 3.3

Independence variances ko add karne deta hai: $Var (S) = Var (X_{1}) + Var (X_{2}) = \frac{35}{12} + \frac{35}{12} = \frac{70}{12} = \frac{35}{6} \approx 5.833.$ Average ke liye, pehle scale karo phir independence use karo: $Var (\overset{ˉ}{X}) = Var (\frac{1}{2} S) = \frac{1}{4} Var (S) = \frac{1}{4} \cdot \frac{35}{6} = \frac{35}{24} \approx 1.458.$ Average kam spread hai ek single roll se — Law of Large Numbers ka seed: kai independent copies ka average variance $1/ n$ ki tarah shrink karta hai.

Level 4 — Synthesis

Goal: kai rules chain karo aur standardise karo.

Recall Solution 4.1

$Z = a X + b$ likho jahan $a = \frac{1}{σ} = \frac{1}{5}$ aur $b = - \frac{μ}{σ} = - \frac{50}{5} = - 10$ . Mean: $E [Z] = \frac{1}{5} E [X] - 10 = \frac{1}{5} (50) - 10 = 10 - 10 = 0.$ Variance: $Var (Z) = (\frac{1}{5})^{2} Var (X) = \frac{1}{25} \cdot 25 = 1.$ Isliye kisi bhi variable ko mean $0$ , variance $1$ ke standard form mein reshape kiya ja sakta hai — Central Limit Theorem mein use hone wala standard form. $μ$ subtract karna use center karta hai; $σ$ se divide karna spread ko exactly $1$ par rescale karta hai.

Recall Solution 4.2

Equal weights. Independence ke saath cross term $0$ hai: $Var (0.5 X + 0.5 Y) = 0. 5^{2} (4) + 0. 5^{2} (16) = 0.25 (4) + 0.25 (16) = 1 + 4 = 5.$ Optimal weight. $W = w X + (1 - w) Y$ ke liye (independent), $Var (W) = w^{2} (4) + (1 - w)^{2} (16) .$ Differentiate karo aur zero set karo (calculus is upward parabola ka bottom dhundhta hai — flat point jahan slope $= 0$ ): $\frac{d}{d w} [4 w^{2} + 16 (1 - w)^{2}] = 8 w - 32 (1 - w) = 0 ⟹ 8 w - 32 + 32 w = 0 ⟹ 40 w = 32 ⟹ w = 0.8.$ Minimum variance: $Var (W) = 0. 8^{2} (4) + 0. 2^{2} (16) = 0.64 (4) + 0.04 (16) = 2.56 + 0.64 = 3.2.$ Dhyan do ki safer asset ( $Var = 4$ ) ko bada weight $0.8$ milta hai — total spread kam karne ke liye tum calmer variable par zyada rely karte ho.

Level 5 — Mastery

Goal: definitions se prove karo; memorised results mein plug-in mat karo.

Recall Solution 5.1

Maano $V = a X + b$ . Linearity se $E [V] = a μ + b$ jahan $μ = E [X]$ . Tab $V - E [V] = (a X + b) - (a μ + b) = a (X - μ) .$ $+ b$ cancel ho jaata hai — yahi wajah hai ki shift spread nahi badal sakta. Ab square karo aur expectation lo: $Var (V) = E [a^{2} (X - μ)^{2}] = a^{2} E [(X - μ)^{2}] = a^{2} Var (X),$ constant $a^{2}$ ko linearity se bahar kheenchte hue. $■$

Recall Solution 5.2

Linearity se expand karo, $c$ ko constant treat karte hue: $M (c) = E [X^{2}] - 2 c E [X] + c^{2} .$ Yeh $c$ mein ek upward parabola hai. Differentiate karo aur zero set karo: $M^{'} (c) = - 2 E [X] + 2 c = 0 ⟹ c = E [X] .$ (Second derivative $M^{''} (c) = 2 > 0$ minimum confirm karta hai.) $c = μ = E [X]$ wapas plug karo: $M (μ) = E [X^{2}] - 2 μ \cdot μ + μ^{2} = E [X^{2}] - μ^{2} = Var (X) .$ Isliye mean squared error ke under $X$ ka sabse best constant predictor mean hai, aur unavoidable leftover error exactly variance hai. $■$

Recall Solution 5.3

Computational formula se, $E [X^{2}] - (E [X])^{2} = Var (X)$ . Variance non-negative quantity $(X - μ)^{2}$ ka average hai, isliye $Var (X) \geq 0$ , jo deta hai $E [X^{2}] \geq (E [X])^{2} .$ Equality tab hold hoti hai jab $Var (X) = 0$ , yaani jab $(X - μ)^{2} = 0$ probability $1$ ke saath — matlab $X$ ek constant hai (bilkul koi randomness nahi). Yeh convex function $g (x) = x^{2}$ ke liye Jensen's Inequality ka sabse simple instance hai: pehle average karna phir square karna, pehle square karna phir average karne se kam nikalta hai. $■$

Recall ladder

Recall Quick self-test (answers chhupao)

Var(3X-2) jab Var(X)=9 ::: 81 Var(X+X) jab Var(X)=5 ::: 20 Var(X-Y), independent, Var(X)=4, Var(Y)=9 ::: 13 Best constant predictor c jo E[(X-c)^2] minimise kare ::: c = E[X] X-mu ko kis quantity se divide karo taaki Var = 1 ho ::: sigma (the SD, variance nahi) E[X^2]-(E[X])^2 ki value ::: Var(X), hamesha >= 0