Worked examples — Linear regression — least squares, inference on coefficients

4.9.22 · D3 · Maths › Probability Theory & Statistics › Linear regression — least squares, inference on coefficients

Shuru karne se pehle, yaad karo woh paanch numbers jinse sab kuch banta hai — ek baar compute karo, baar baar use karo. Poore note mein $n$ number of data points hai (kitne $(x_i,y_i)$ pairs hain); chhota subscript $i$ bas yeh batata hai ki hum kaunsa point le rahe hain, jo $i=1,2,\dots ,n$ tak jaata hai.

Recall Woh paanch quantities jo tum hamesha pehle compute karte ho

$\bar{x}$ (mean of $x$) ::: tumhare inputs ka "centre" horizontal direction mein $\bar{y}$ (mean of $y$) ::: tumhare outputs ka "centre" vertical direction mein $S_{xx}=\sum (x_i-\bar{x}{)}^2$ ::: $x$'s kitne spread out hain (hamesha $\ge 0$) $S_{xy}=\sum (x_i-\bar{x})(y_i-\bar{y})$ ::: $x$ aur $y$ saath mein kaise move karte hain (koi bhi sign ho sakta hai) $S_{yy}=\sum (y_i-\bar{y}{)}^2$ ::: $y$'s ka total spread (yahi SST hai)

Yahan $\sum$ (capital Greek "sigma") ka matlab hai "saare $n$ data points par add karo," aur kisi letter ke upar bar jaise $\bar{x}$ ka matlab hai "us letter ka average."

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Scenario matrix

Har simple-regression problem exactly inhi cells mein se ek mein hogi. Last column us example ka naam deta hai jo use best explain karta hai.

#	Case class	Kya special hai	Dhyan rakho	Example
A	Positive slope	$S_{xy}>0$, points upar jaate hain	ordinary case	Ex 1
B	Negative slope	$S_{xy}<0$, points neeche jaate hain	${\hat{\beta }}_1$ ka sign	Ex 2
C	Zero slope	$S_{xy}=0$ exactly	line flat hai, ${\hat{\beta }}_1=0$, $R^2=0$	Ex 3
D	Degenerate $x$	saare $x_i$ equal hain, $S_{xx}=0$	division by zero — koi line nahi	Ex 4
E	Perfect fit	points exactly ek line par, SSE $=0$	$R^2=1$, SE $=0$, $t=\infty$	Ex 5
F	Constant response	saare $y_i$ equal hain, $S_{yy}=0$	$R^2=0/0$ undefined	Ex 6
G	Real word problem	units, prediction, interpretation	har number par units rakho	Ex 7
H	Inference / significance	kya slope real hai?	$t$ vs critical value, CI	Ex 8
I	Exam twist: rescale $x$	$x$ ki units badlo	${\hat{\beta }}_1$, SE, $R^2$ kaise change hote hain	Ex 9

Ab hum inhe order mein chalte hain. Cells A–C ek picture share karte hain taaki tum dekh sako ki $S_{xy}$ ka sign tilt ko kaise flip karta hai.

Figure s01 — wohi paanch $x$-values teen alag $y$-patterns ke saath. Left (magenta, Cell A): points chadhte hain, fitted navy line upar tilt karti hai. Middle (violet, Cell B): points girte hain, line neeche tilt karti hai. Right (orange, Cell C): points balance out karte hain, line flat hai. Teeno mein star $(\bar{x},\bar{y})$ mark karta hai — notice karo ki har line exactly usse guzarti hai.

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Example 1 — Cell A: positive slope

Step 1 — data ko centre karo. $\bar{x}=\frac{1+2+3+4+5}{5}=3$, $\bar{y}=\frac{1+3+2+5+4}{5}=3$. Yeh step kyun? Poora formula measure karta hai ki points apne centre se kitna deviate karte hain; line ko $(\bar{x},\bar{y})$ se guzarna padta hai, toh hum pehle wahan anchor karte hain.

Step 2 — $x$ ka spread. $S_{xx}=(-2{)}^2+(-1{)}^2+0^2+1^2+2^2=4+1+0+1+4=10.$ Kyun? Yeh "lever length" hai jo baad mein slope ko pin down karega.

Step 3 — co-movement. $y$ ke deviations: $(-2,0,-1,2,1)$. $S_{xy}=(-2)(-2)+(-1)(0)+(0)(-1)+(1)(2)+(2)(1)=4+0+0+2+2=8.$ Kyun? $S_{xy}$ har $x$-deviation ko uske $y$-deviation ke saath pair karta hai. Dono positive ya dono negative → positive product → upward trend.

Step 4 — slope aur intercept.

\hat\beta_0=\bar y-\hat\beta_1\bar x=3-0.8(3)=0.6.$$ Line: $\hat y=0.6+0.8x$. > **Verify:** $(\bar x,\bar y)=(3,3)$ se guzarti hai? $0.6+0.8(3)=0.6+2.4=3$ ✓. Slope $0.8>0$ aur 1 se kam, forecast se match karta hai. --- ## Example 2 — Cell B: negative slope > [!example] Statement > Same $x=(1,2,3,4,5)$ lekin ab $y=(5,4,4,2,1)$. Line fit karo. > > **Forecast:** Dots left-to-right girte hain, toh $\hat\beta_1<0$. Intercept (value at $x=0$, left mein door) $\bar y$ se *upar* hona chahiye. **Step 1 — centres.** $\bar x=3$, $\bar y=\dfrac{5+4+4+2+1}{5}=\dfrac{16}{5}=3.2$. **Step 2 — $S_{xx}=10$** (same $x$ pehle jaise — reuse karo). *Kyun reuse karo?* $S_{xx}$ sirf $x$ par depend karta hai. **Step 3 — co-movement.** $y$-deviations: $(1.8,0.8,0.8,-1.2,-2.2)$. $$S_{xy}=(-2)(1.8)+(-1)(0.8)+0(0.8)+1(-1.2)+2(-2.2)=-3.6-0.8+0-1.2-4.4=-10.$$ *Negative kyun?* Left ke points centre se upar hain, right ke points neeche — opposite signs multiply karke negative dete hain. Woh negative $S_{xy}$ *hi* downward tilt hai. **Step 4 — slope aur intercept.** $$\hat\beta_1=\frac{-10}{10}=-1,\qquad \hat\beta_0=3.2-(-1)(3)=6.2.$$ Line: $\hat y=6.2-x$. > **Verify:** $(3,3.2)$ se guzarti hai? $6.2-3=3.2$ ✓. Slope $-1<0$; intercept $6.2>\bar y=3.2$ — dono forecast se match karte hain. --- ## Example 3 — Cell C: exactly zero slope > [!example] Statement > $x=(1,2,3,4,5)$, $y=(2,4,3,4,2)$ — ek symmetric hump. Line fit karo aur $R^2$ nikalo. > > **Forecast:** Pattern pehle chadhta hai phir girta hai, koi overall tilt nahi; "best line" $y$ ke mean par **flat** honi chahiye, giving $\hat\beta_1=0$ aur $R^2=0$ (flat line koi linear variation explain nahi karti). **Step 1 — centres.** $\bar x=3$, $\bar y=\dfrac{2+4+3+4+2}{5}=3$. **Step 2 — $S_{xx}=10$**, aur $y$-deviations $(-1,1,0,1,-1)$. **Step 3 — co-movement.** $$S_{xy}=(-2)(-1)+(-1)(1)+(0)(0)+(1)(1)+(2)(-1)=2-1+0+1-2=0.$$ *Zero kyun?* Left side ($x<\bar x$) aur right side ($x>\bar x$) equal aur opposite amounts contribute karte hain — exactly cancel ho jaate hain. Koi bhi linear co-movement nahi. **Step 4 — estimates.** $\hat\beta_1=0/10=0$, $\hat\beta_0=3-0=3$. Line: $\hat y=3$ (flat). **Step 5 — goodness of fit.** Jab $\hat y_i=\bar y$ sab $i$ ke liye, har fitted value mean ke barabar hai, toh explained sum $\text{SSR}=\sum(\hat y_i-\bar y)^2=0$, jis se $$R^2=\frac{\text{SSR}}{\text{SST}}=0.$$ > **Verify:** $\text{SST}=(-1)^2+1^2+0^2+1^2+(-1)^2=4$. $\text{SSE}=\sum(y_i-3)^2=4$ (line har jagah $3$ predict karti hai). Phir $R^2=1-\text{SSE}/\text{SST}=1-4/4=0$ ✓. Flat line variation ka **0\%** explain karti hai — Cells A aur B ke beech ki boundary. --- ## Example 4 — Cell D: degenerate input, koi line exist nahi karti > [!example] Statement > $x=(2,2,2,2)$, $y=(1,3,2,4)$. Line fit karne ki koshish karo. > > **Forecast:** Saare $x$'s ek vertical line par stack hain. Measure karne ke liye koi "left-to-right" tilt nahi hai — tum us stack se infinitely many lines draw kar sakte ho. Toh slope **undefined** hona chahiye. **Step 1 — $x$ ka spread.** $\bar x=2$, toh har deviation $x_i-\bar x=0$ aur $$S_{xx}=0^2+0^2+0^2+0^2=0.$$ **Step 2 — slope nikalne ki koshish.** $$\hat\beta_1=\frac{S_{xy}}{S_{xx}}=\frac{S_{xy}}{0}\quad\Rightarrow\quad \textbf{undefined}.$$ *Yeh step kyun matter karta hai:* $0$ se divide karna huge number nahi hota — yeh *koi answer nahi* hota. Geometrically "lever" ki zero length hai, toh slope ko aim hi nahi kiya ja sakta. > [!mistake] "$S_{xx}=0$ toh bas slope ko 0 set kar do." > *Kyun sahi lagta hai:* kisi chhoti cheez se divide karne par usually definite value milti hai. **Fix:** yahan numerator ka meaning ($x$-weighted co-movement) bhi collapse ho jaata hai — genuinely *koi* least-squares line nahi hai. Tum sirf ek constant $\hat\beta_0=\bar y=2.5$ estimate kar sakte ho. Statistical software typically yeh flag karta hai — package ke hisaab se tumhe warning, `NA`/`NaN` slope, ya outright error mil sakta hai — lekin *kabhi* trustworthy finite slope nahi. > **Verify:** $S_{xx}=0$, toh $\hat\beta_1=S_{xy}/S_{xx}$ ka zero denominator hai — undefined ✓. Neeche figure mein vertical stack dekho. ![[deepdives/dd-maths-4.9.22-d3-s02.png]] *Figure s02 — Cell D. Saare chaar points $x=2$ share karte hain, toh woh ek vertical dashed line par stack hote hain. Star $(\bar x,\bar y)=(2,2.5)$ mark karta hai. Star se guzarne wali teen colored dashed lines sarabraar "fit" karti hain — koi horizontal spread nahi ($S_{xx}=0$) toh koi ek tilt prefer karne ka koi tarika nahi, isliye slope undefined hai.* --- ## Example 5 — Cell E: perfect fit, limiting case > [!example] Statement > $x=(1,2,3,4)$, $y=(3,5,7,9)$ — har point exactly ek line par baitha hai. Fit karo aur $R^2$, SSE, aur slope ka standard error nikalo. > > **Forecast:** Agar points *exactly* linear hain toh koi leftover noise nahi. Expect karo SSE $=0$, $R^2=1$, aur $\text{SE}(\hat\beta_1)=0$ (koi uncertainty nahi). **Step 1 — centres.** $\bar x=2.5$, $\bar y=\dfrac{3+5+7+9}{4}=6$. **Step 2 — sums.** $x$-deviations $(-1.5,-0.5,0.5,1.5)$, $y$-deviations $(-3,-1,1,3)$. $$S_{xx}=2.25+0.25+0.25+2.25=5,\qquad S_{xy}=4.5+0.5+0.5+4.5=10.$$ **Step 3 — line.** $\hat\beta_1=10/5=2$, $\hat\beta_0=6-2(2.5)=1$. Line: $\hat y=1+2x$. **Step 4 — residuals.** Predictions: $3,5,7,9$ — data ke identical. Toh har $e_i=0$, jis se $\text{SSE}=0$. **Step 5 — goodness aur precision.** $$R^2=1-\frac{\text{SSE}}{\text{SST}}=1-\frac{0}{20}=1,\qquad s^2=\frac{\text{SSE}}{n-2}=\frac{0}{2}=0,\qquad \text{SE}(\hat\beta_1)=\frac{s}{\sqrt{S_{xx}}}=0.$$ *SE zero kyun hai:* koi scatter nahi, toh har resampled dataset *same* line deta hai — zero wobble. $t$-statistic $\hat\beta_1/\text{SE}$ hoga $2/0=\infty$: infinitely significant, jaise hona chahiye. > **Verify:** SST $=9+1+1+9=20$; SSE $=0$; $R^2=1$ ✓; $s^2=0$; SE $=0$ ✓. Yeh Cell C ($R^2=0$) ka opposite extreme hai. --- ## Example 6 — Cell F: constant response (saare $y$ equal) > [!example] Statement > $x=(1,2,3,4,5)$, $y=(7,7,7,7,7)$ — output kabhi nahi badalta. Line fit karo aur $R^2$ nikalne ki koshish karo. > > **Forecast:** $y$ flat hai, toh slope ke explain karne ke liye *kuch* nahi aur explain hone ke liye *kuch* nahi. Expect karo $\hat\beta_1=0$, lekin $R^2$ **undefined** — tum zero total variation ka fraction nahi le sakte. **Step 1 — centres.** $\bar x=3$, $\bar y=7$. **Step 2 — sums.** $y$-deviations sab $0$ hain, toh $$S_{xy}=\sum(x_i-\bar x)(y_i-\bar y)=0,\qquad S_{yy}=\text{SST}=\sum(y_i-7)^2=0.$$ $S_{xx}=10$ ($x$'s ab bhi spread hain). **Step 3 — slope aur line.** $\hat\beta_1=0/10=0$, $\hat\beta_0=7-0=7$. Fitted line $\hat y=7$ hai, jo har data point ko *exactly* reproduce karti hai, toh $\text{SSE}=0$ bhi. **Step 4 — $R^2$ nikalne ki koshish.** $$R^2=\frac{\text{SSR}}{\text{SST}}=\frac{0}{0}\quad\Rightarrow\quad \textbf{undefined}.$$ *Yeh step kyun matter karta hai:* $R^2$ ka jawab hai "line ne $y$ ki variation ka kitna fraction capture kiya?" — lekin yahan $y$ ki **koi** variation nahi, toh sawaal khud hi empty hai ($0/0$). > [!mistake] "$R^2=0$ jab $y$ constant ho, kyunki line kuch explain nahi karti." > *Kyun sahi lagta hai:* SSR $=0$, aur $0/\text{kuch}=0$. **Fix:** denominator SST *bhi* $0$ hai, toh $R^2=0/0$ genuinely undefined hai, $0$ nahi. Cell C se compare karo, jahan SST $>0$ tha aur clean $R^2=0$ mila. Inference yahan bhi meaningless hai: $s^2=0$ toh har SE $=0$ aur $t$-ratio $0/0$ hai. > **Verify:** $S_{yy}=0$ aur $\text{SSR}=0$, toh $R^2=\text{SSR}/\text{SST}=0/0$ — undefined ✓. Slope $\hat\beta_1=0$ well-defined hai kyunki $S_{xx}=10\neq0$. --- ## Example 7 — Cell G: real-world word problem with units > [!example] Statement > Ek café daily high temperature $x$ (°C mein) aur becha hua iced coffee $y$ cups record karta hai: > $x=(20,22,24,26,28)$, $y=(30,35,40,48,52)$ cups. Line fit karo aur **25 °C par sales predict karo**. > > **Forecast:** Zyada garmi → zyada iced coffee, toh $\hat\beta_1>0$. Roughly sales 8 °C range mein ~22 cups badhti hain, toh slope around $22/8\approx 2.7$ cups per °C hoga. **Step 1 — centres.** $\bar x=24$ °C, $\bar y=\dfrac{30+35+40+48+52}{5}=41$ cups. **Step 2 — sums.** $x$-deviations $(-4,-2,0,2,4)$, $y$-deviations $(-11,-6,-1,7,11)$. $$S_{xx}=16+4+0+4+16=40\ (\text{°C}^2),$$ $$S_{xy}=(-4)(-11)+(-2)(-6)+0(-1)+2(7)+4(11)=44+12+0+14+44=114\ (\text{°C·cups}).$$ **Step 3 — slope with units.** $$\hat\beta_1=\frac{114\ \text{°C·cups}}{40\ \text{°C}^2}=2.85\ \frac{\text{cups}}{\text{°C}}.$$ *Units kyun track karo?* Slope ka matlab hai "extra cups per extra degree." Units sahi rakhna sanity check hai ki tumne sahi pair divide kiya. **Step 4 — intercept aur prediction.** $$\hat\beta_0=41-2.85(24)=41-68.4=-27.4\ \text{cups}.$$ $x=25$ °C par: $\hat y=-27.4+2.85(25)=-27.4+71.25=43.85\approx 44$ cups. > [!mistake] "Intercept $=-27.4$ matlab café 0 °C par negative coffee bechta hai." > *Kyun sahi lagta hai:* $\hat\beta_0$ value at $x=0$ hai. **Fix:** $x=0$ °C data range (20–28 °C) se bahut door hai. Wahan line extrapolate karna meaningless hai — intercept sirf ek fitting anchor hai, freezing par real forecast nahi. > **Verify:** Slope $2.85$ cups/°C ~2.7 forecast se match karta hai. $\bar x=24$ par prediction $\bar y$ ke barabar honi chahiye: $-27.4+2.85(24)=41$ ✓. $25$ °C par, $43.85$ cups ✓ — mean se thoda upar, centre se ek degree upar expected hai. --- ## Example 8 — Cell H: kya slope real hai? (inference) > [!example] Statement > Café data (Ex 7) reuse karo. $H_0:\beta_1=0$ (temperature ka koi effect nahi) ko 5\% level par test karo, aur slope ke liye 95\% confidence interval do. > > **Forecast:** Upward trend bahut clean lag raha tha, toh expect karo large $t$-statistic aur CI jo 0 se kaafi upar ho — hum "no effect" ko **reject** karna chahiye. **Step 1 — residuals.** $\hat y=-27.4+2.85x$ ke saath, $x=20,22,24,26,28$ par fitted values hain $29.6,\ 35.3,\ 41.0,\ 46.7,\ 52.4$. Data $y=(30,35,40,48,52)$ se subtract karne par residuals milte hain $$e=y-\hat y=(0.4,\ -0.3,\ -1.0,\ 1.3,\ -0.4).$$ *Residuals kyun?* Woh unseen noise tak hamari ek maatra window hain, aur wahi noise slope mein saari uncertainty drive karti hai. **Step 2 — residual variance $s^2$ ($\sigma^2$ ka hamara estimate).** $$\text{SSE}=0.4^2+(-0.3)^2+(-1.0)^2+1.3^2+(-0.4)^2=0.16+0.09+1.0+1.69+0.16=3.1.$$ $$s^2=\frac{\text{SSE}}{n-2}=\frac{3.1}{5-2}=\frac{3.1}{3}=1.0333,\qquad s=\sqrt{1.0333}=1.0165.$$ *$n$ ki jagah $n-2$ se divide kyun karo?* Hamare paas $n=5$ points hain lekin humne 2 unhe $\hat\beta_0$ aur $\hat\beta_1$ pin karne mein spend kar diye; sirf $n-2=3$ independent pieces of noise information bachi hain. **Step 3 — slope ka standard error.** $$\text{SE}(\hat\beta_1)=\frac{s}{\sqrt{S_{xx}}}=\frac{1.0165}{\sqrt{40}}=\frac{1.0165}{6.3246}=0.16072.$$ *Yeh formula kyun?* Slope noisy $y_i$ ka weighted sum hai; unka variance propagate karne par $\text{Var}(\hat\beta_1)=\sigma^2/S_{xx}$ milta hai, aur hum unknown $\sigma$ ki jagah $s$ plug in karte hain. **Step 4 — $t$-statistic.** $$t=\frac{\hat\beta_1-0}{\text{SE}(\hat\beta_1)}=\frac{2.85}{0.16072}=17.73\quad\text{on }n-2=3\text{ df}.$$ *$t$-distribution se kyun compare karo?* Kyunki humne $\sigma$ us chhote sample se hi estimate kiya, extra uncertainty tails ko mota karti hai — exactly yahi [[t-distribution]] $3$ degrees of freedom ke saath encode karta hai (dekho [[Hypothesis Testing]]). **Step 5 — decision aur 95\% confidence interval.** Critical value hai $t_{3,\,0.975}=3.182$. Kyunki $17.73>3.182$, hum $H_0$ ko **reject** karte hain: effect real hai. Interval hai $$\hat\beta_1\pm t_{3,0.975}\cdot\text{SE}(\hat\beta_1)=2.85\pm 3.182(0.16072)=2.85\pm 0.5114=[2.34,\ 3.36]\ \text{cups/°C}.$$ > **Verify:** $\text{SSE}=3.1$; $s^2=3.1/3\approx1.0333$; $\text{SE}\approx0.1607$; $t\approx17.7$ ✓. > CI $[2.34,3.36]$ $0$ ko exclude karta hai, rejection ke saath consistent hai — exactly forecast se match karta hai ki yeh clean trend significant hai. --- ## Example 9 — Cell I: exam twist — $x$-axis rescale karo > [!example] Statement > Café problem (Ex 7–8) mein, ek intern temperature **°F** mein record karta hai °C ki jagah, toh har $x_i$ ban jaata hai $x_i'=1.8x_i+32$. **Scratch se refit kiye bina**, $\hat\beta_1$, $R^2$, aur $\text{SE}(\hat\beta_1)$ kaise change hote hain? > > **Forecast:** Har degree °F ek °C se *chhota* step hai, toh "cups per degree" factor $1.8$ se **shrink** hona chahiye. Points ka *pattern* unchanged hai, toh $R^2$ same rehna chahiye. SE slope ke same proportion mein shrink hona chahiye. **Step 1 — sums kaise transform hoti hain.** Linear rescale $x'=a x+b$ (yahan $a=1.8$, $b=32$) deviations ko shift aur stretch karta hai: $x_i'-\bar{x'}=a(x_i-\bar x)$. Isliye $$S_{x'x'}=a^2 S_{xx},\qquad S_{x'y}=a\,S_{xy}.$$ *Yeh step kyun?* Additive shift $b$ har deviation mein cancel ho jaata hai; sirf multiplier $a$ bachta hai, aur woh $S_{x'x'}$ mein *do baar* bachta hai (do $x$-factors) lekin $S_{x'y}$ mein *ek baar*. **Step 2 — new slope.** $$\hat\beta_1'=\frac{S_{x'y}}{S_{x'x'}}=\frac{a\,S_{xy}}{a^2 S_{xx}}=\frac{1}{a}\hat\beta_1 =\frac{2.85}{1.8}=1.5833\ \frac{\text{cups}}{\text{°F}}.$$ **Step 3 — $R^2$ invariant hai.** $R^2=r^2$ jahan [[Correlation coefficient|correlation]] $r=\dfrac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}$. Rescale ke under $r'=\dfrac{aS_{xy}}{\sqrt{a^2S_{xx}\,S_{yy}}}=\dfrac{a}{|a|}\,r=r$ (kyunki $a=1.8>0$). Toh $R^2$ **unchanged** hai. *Kyun?* Correlation units ignore karta hai — yeh pattern ke baare mein hai, scale ke nahi. **Step 4 — new standard error.** SSE aur $s$ sirf $y$ aur residuals par depend karte hain, jo change nahi hote (fitted values identical hain, bas ek stretched axis par relabelled). Toh $s$ unchanged hai, aur $$\text{SE}(\hat\beta_1')=\frac{s}{\sqrt{S_{x'x'}}}=\frac{s}{a\sqrt{S_{xx}}}=\frac{1}{a}\text{SE}(\hat\beta_1) =\frac{0.16072}{1.8}=0.08929.$$ **Step 5 — $t$-statistic invariant hai.** $$t'=\frac{\hat\beta_1'}{\text{SE}(\hat\beta_1')}=\frac{\hat\beta_1/a}{\text{SE}(\hat\beta_1)/a} =\frac{\hat\beta_1}{\text{SE}(\hat\beta_1)}=t=17.73.$$ *Yeh beautiful kyun hai:* units badalna nahi badal sakta ki effect significant hai ya nahi. Slope number move hua, lekin *evidence* nahi. > **Verify:** $\hat\beta_1'=2.85/1.8\approx1.583$ ✓; $\text{SE}=0.1607/1.8\approx0.0893$ ✓; $t$ $\approx17.7$ par unchanged ✓; $R^2$ identical. Slope aur SE dono $1/1.8$ se scale hue, exactly forecast se match karta hai. --- ## Har cell ne apne tools kahan use kiye ```mermaid flowchart TD A["Compute Sxx and Sxy"] --> B{"Is Sxx = 0?"} B -->|yes| C["Degenerate: no line, Cell D"] B -->|no| D["Slope = Sxy over Sxx"] D --> E{"Sign of Sxy"} E -->|positive| F["Up trend, Cell A"] E -->|zero| G["Flat line, R2 = 0, Cell C"] E -->|negative| H["Down trend, Cell B"] D --> I{"Is SST = 0?"} I -->|yes| J["Constant y, R2 undefined, Cell F"] I -->|no| K{"Is SSE = 0?"} K -->|yes| L["Perfect fit, R2 = 1, Cell E"] K -->|no| M["Inference: t and CI, Cell H"] ``` --- > [!recall]- Jaldi self-test > Agar saare $x_i$ equal hain, toh $\hat\beta_1$ kya hai? ::: Undefined — $S_{xx}=0$, division by zero (Cell D). > Agar saare $y_i$ equal hain, toh $R^2$ kya hai? ::: Undefined — $\text{SST}=0$ se $0/0$ milta hai (Cell F). > $S_{xy}=0$ exactly. $\hat\beta_1$ aur $R^2$ kya hain? ::: Dono $0$; best line $\bar y$ par flat hai (Cell C). > Points exactly ek line par hain. Slope ka SE kya hai? ::: $0$ — SSE $=0$ toh $s=0$ (Cell E). > Tum $x$ ko cm se mm mein badle ($x'=10x$). $t$-statistic ka kya hota hai? ::: Kuch nahi — $t$ unit-invariant hai (Cell I). > Positive $S_{xy}$ batata hai ki slope…? ::: Positive hai — $x$ aur $y$ saath upar co-move karte hain (Cell A). > [!mnemonic] Poora matrix ek saansi mein > **"$S_{xy}$ ka sign tilt set karta hai; $S_{xx}=0$ fit ko khatam karta hai; $S_{yy}=0$ $R^2$ ko khatam karta hai; SSE $=0$ use perfect banata hai; units number ko move karte hain, $t$ ko kabhi nahi."** Dekho: [[Correlation coefficient]] ($R^2$ drive karta hai), [[Covariance]] (un-normalised $S_{xy}$), [[t-distribution]] aur [[Hypothesis Testing]] (Cell H), [[Gauss–Markov Theorem]] (kyun least squares best hai), [[Maximum Likelihood Estimation]] (kyun squared errors), aur [[Multiple Regression]] (same story zyada $x$'s ke saath).