Expected value, variance, standard deviation — properties: Visual walkthrough

4.9.4 · D2 · Maths › Probability Theory & Statistics › Expected value, variance, standard deviation — properties

Step 0 — Pehle do zaroori ideas samajhni hain

Kisi bhi formula se pehle, hume do ideas ek picture se anchor karni hain.

Figure padhna. Har coral blob ek probability lump hai jo apni value par baitha hai. Lavender triangle knife edge hai — yeh exactly $μ$ ke neeche baitha hai. Term by term:

$x_{i}$ — lump kahan baitha hai (ruler par position),
$p_{i}$ — lump kitna bhaari hai (uska probability share),
$x_{i} p_{i}$ — us lump ka origin ke baare mein "turning effort" (torque),
$\sum_{i} x_{i} p_{i}$ — total turning effort, jise balance point $μ$ ko cancel karna padta hai.

Step 1 — "Spread" ka matlab kya hai?

KYA. Hum EK number chahte hain jo bataye ki lumps balance point se ausat kitni door hain.

KYUN. Do experiments ka exact same mean $μ$ ho sakta hai lekin feel bilkul alag hota hai: ek tight cluster, ek wild scatter. Mean akela yeh nahi dekh sakta. Hume ek spread number chahiye.

PICTURE. Har lump ka signed distance $d_{i} = x_{i} - μ$ dekho — knife edge se lump tak ek arrow.

$d_{i} = x_{i} - μ$ — balance point se lump $i$ tak ka arrow. Mint = lump dayi taraf hai ( $d_{i} > 0$ ); coral = bayi taraf ( $d_{i} < 0$ ).

Pehla tempting idea: bas in arrows ko average karo, $E [d] = E [X - μ]$ . Agla step dikhata hai kyun yeh fail hota hai.

Step 2 — Raw distances kyun bekaar hain (unka sum zero hota hai)

KYA. Signed arrows $d_{i} = x_{i} - μ$ ka average compute karo.

KYUN. Dikhane ke liye ki yeh hamesha $0$ par collapse karta hai — isliye hume ise theek karna padega.

$E [X - μ] = = μ E [X] - a constant μ = μ - μ = 0.$

Term by term: $E [X]$ khud balance point $μ$ hai; $μ$ ek fixed number hai isliye uska average bas $μ$ hi hai. Dono cancel ho jaate hain. Yeh coincidence nahi hai — mean ko hi wo point define kiya jaata hai jahan left-pull aur right-pull exactly balance karte hain.

PICTURE. Mint (dayi) arrows aur coral (bayi) arrows ka total "length-times-weight" barabar hai. Ye ek doosre ko annihilate kar dete hain.

Step 3 — Fix: har arrow ko square karo

KYA. Har signed distance $d_{i}$ ko uske square $d_{i}^{2} = (x_{i} - μ)^{2}$ se replace karo, phir average karo.

Squaring KYUN, aur kyun nahi, say, absolute value $∣ d_{i} ∣$ ? Teen reasons, sab real:

Squaring sign ko khatam karta hai (negative ko square karo to positive milta hai) — isliye kuch cancel nahi hota.
Yeh bade misses ko zyada punish karta hai — do baar door baitha lump chaar baar spread contribute karta hai.
Yeh smooth hai (ek parabola), isliye baad ki algebra aur calculus behave karti hai; $∣ d ∣$ mein ek sharp corner hai.

Isse variance define hota hai: $Var (X) = E [(X - μ)^{2}] .$

$(X - μ)$ — signed arrow,
$(\cdot)^{2}$ — sign-remover jo bade arrows ko aur bhi magnify karta hai,
$E [\dots]$ — saare lumps par average, $p_{i}$ se weight karke.

PICTURE. Har arrow $d_{i}$ us par bana ek square ka area ban jaata hai. Left aur right dono genuine (positive) area dete hain.

Recall Kyun nahi bas

$E [∣ X - μ ∣]$ ? Yeh ek valid spread measure hai (the "mean absolute deviation"), lekin $0$ par iska corner ise non-differentiable banata hai aur iska algebra ugly hai. Squaring se woh clean expandable formula milta hai jo hum aage chahte hain. ::: Because $x^{2}$ is smooth and expands algebraically, while $∣ x ∣$ has a non-differentiable corner.

Step 4 — Square ko expand karo (key algebraic move)

KYA. Expectation ke andar $(X - μ)^{2}$ ko multiply out karo.

KYUN. Definition $E [(X - μ)^{2}]$ mein $μ$ pehle se pata hona chahiye — ek two-pass kaam. Expand karne se hum ise aisi pieces mein reorganise kar sakte hain jo data ke ek pass mein compute ho sakti hain.

(X - μ)^{2} = X^{2} - 2 μ X + μ^{2} .

$X^{2}$ — value ka square (outcome ke saath vary karta hai),
$- 2 μ X$ — ek cross term; note karo $μ$ yahan ek fixed constant hai,
$μ^{2}$ — ek pure constant (balance point ka square).

PICTURE. Classic square-of-a-sum: side $∣ X - μ ∣$ ka ek bada square ek $X^{2}$ block, ek $μ^{2}$ block, aur do identical $μ X$ rectangles mein decompose hota hai.

Step 5 — Linearity use karke har piece ka average nikalo

KYA. Expanded expression ke har piece ka $E [\dots]$ lo.

Linearity yahan kyun legal hai: ek sum ka average, averages ka sum hota hai, aur ek constant average se bahar aa jaata hai. (Yeh wahi linearity rule hai jo parent ne prove kiya tha.)

Var (X) = E [X^{2}] - 2 μ E [X] + μ^{2} .

$E [X^{2}]$ — squares ka average (LOTUS chahiye: average $g (x) = x^{2}$ ),
$- 2 μ E [X]$ — constant $2 μ$ average ke aage aagaya,
$μ^{2}$ — ek constant ka average woh constant khud hota hai.

Ab yeh fact substitute karo ki $E [X] = μ$ :

Var (X) = E [X^{2}] - 2 μ \cdot μ + μ^{2} = E [X^{2}] - 2 μ^{2} + μ^{2} = E [X^{2}] - μ^{2} .

Beech ke dono terms $- 2 μ^{2} + μ^{2}$ collapse hokar $- μ^{2}$ ban jaate hain. $μ = E [X]$ wapas likhke:

PICTURE. Do competing bars: lamba $E [X^{2}]$ bar, minus chhota $(E [X])^{2}$ bar; bacha hua slab variance hai.

Step 6 — Answer kabhi negative kyun nahi hota

KYA. Confirm karo ki $Var (X) \geq 0$ hamesha hota hai — yaani $E [X^{2}] \geq (E [X])^{2}$ .

Aise hona KYUN zaroori hai. Variance ko Step 3 mein squares $(X - μ)^{2}$ ke average ke roop mein define kiya gaya tha. Squares kabhi negative nahi hote, aur non-negative cheezein average karne par non-negative rehta hai. Isliye lamba bar kabhi chhote bar se chhota nahi ho sakta.

\quad\Longrightarrow\quad \mathbb{E}[X^2]\ge(\mathbb{E}[X])^2.$$ Yeh convex function $x^2$ ke liye [[Jensen's Inequality|Jensen's inequality]] exactly hai: *squares ka average, average ke square se bada hota hai.* **PICTURE.** Ek convex parabola $y=x^2$: chord (jo $\mathbb{E}[X^2]$ represent karta hai) mean par curve point (jo $(\mathbb{E}[X])^2$ hai) ke **upar** baitha hai. Unke beech ka gap *variance hai*. --- ## Step 7 — Edge cases (kuch bhi chhupa nahi) **Case A — ek certain outcome (koi randomness nahi).** Agar $X$ hamesha ek value $c$ barabar ho (isliye wahan $p=1$ hai), to har lump balance point par baitha hai, isliye har arrow $d_i=0$ hai. Variance $=\mathbb{E}[0]=0$. Lamba aur chhota bar barabar hain. *Ek sure cheez ka spread zero hota hai.* **Case B — $b$ se shift karna.** Har value slide karo: $X\to X+b$. Har lump aur knife edge saath chalte hain, isliye saare arrows $d_i$ **unchanged** rehte hain. Isliye $\operatorname{Var}(X+b)=\operatorname{Var}(X)$. $b$ gayab ho jaata hai. **Case C — $a$ se scale karna.** Stretch karo: $X\to aX$. Har arrow $a$ guna lamba ho jaata hai, isliye uska square $a^2$ guna bada ho jaata hai: $\operatorname{Var}(aX)=a^2\operatorname{Var}(X)$. Factor **square** hota hai kyunki variance squared units mein rehta hai — aur yeh $a^2\ge 0$ hai chahe $a$ negative ho (ek mirror-flip spread ko nahi ghataata). - Left panel: certain outcome — ek lump, zero arrows, $\operatorname{Var}=0$. - Middle panel: shift — picture slide karti hai, arrows same hain, $\operatorname{Var}$ same hai. - Right panel: scale — picture stretch hoti hai, arrows $\times a$, variance $\times a^2$. > [!example] Sanity check — fair die > $\mathbb{E}[X]=3.5$, $\ \mathbb{E}[X^2]=\dfrac{91}{6}\approx 15.1\overline{6}$. > $$\operatorname{Var}(X)=\tfrac{91}{6}-(3.5)^2=15.1\overline{6}-12.25=2.91\overline{6}.$$ > Positive ✓ (Step 6). Aur $Y=10X+5$: shift kuch nahi karta, scale square hota hai → $\operatorname{Var}(Y)=10^2\cdot 2.91\overline{6}=291.\overline{6}$ (Step 7 C). $\ \operatorname{SD}(Y)=|10|\cdot\sqrt{2.91\overline{6}}\approx 17.08$. --- ## Ek-picture summary Sab kuch ek canvas par: ruler par lumps → balance point se arrows → un arrows ke squares → unka average variance hai → jo tall bar $\mathbb{E}[X^2]$ minus short bar $(\mathbb{E}[X])^2$ ke barabar hai → aur woh gap kabhi negative nahi ho sakta. > [!recall]- Feynman retelling (ise zaur se bolo) > Kuch clay lumps ek ruler par bikhaao; weights probabilities hain. Dekho yeh kahan balance karta hai — woh **mean** hai. Ab main jaanna chahta hoon ki clay kitni spread out hai. Main balance point se har lump tak ek arrow kheenchta hoon. Agar main bas un arrows ko average karun to mujhe zero milta hai, kyunki left-pull aur right-pull cancel ho jaate hain — "balance point" ka yehi matlab hai! Isliye main har arrow par ek chhota square banata hoon (squaring sign ko mita deta hai aur door ke lumps ko extra punish karta hai), aur main un squares ko average karta hoon. Woh average **variance** hai. Thodi si algebra se, wahi average saaf saaf "squared values ka average" minus "average value ka square" — tall bar minus short bar — mein split ho jaata hai. Kyunki main squares average karne se shuru hua tha, answer kabhi zero se neeche nahi ja sakta: tall bar hamesha jeetta hai. Poora ruler slide karo aur kuch nahi badalta; use ek factor se stretch karo aur spread us factor ke *square* se badhta hai. > [!mnemonic] Formula yaad rakho > **"Square ka mean, minus mean ka square."** Order matter karta hai — inhe swap karo aur ek negative number milega, jo spread kabhi nahi ho sakta. --- **Prerequisites & neighbours:** [[Probability Distributions]] · [[Covariance and Correlation]] · [[Jensen's Inequality]] · [[Bernoulli and Binomial Distributions]] · [[Law of Large Numbers]] · [[Central Limit Theorem]] · [[Moment Generating Functions]]