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Another approach

$$f_{X|Y}(x,y)=\frac{f_{X,Y}(x,y)}{f_{Y}(y)}\tag 1$$ Set $$u=\frac{x-\mu_X}{\sigma_X}$$ and $$v=\frac{y-\mu_Y}{\sigma_Y}$$ we have $$f_{X|Y}(x,y)=\frac{\frac{1}{2\pi\sigma_X\sigma_Y\sqrt{1-\rho^2}}\exp\left(-\frac{u^2-2\rho uv+v^2}{2(1-\rho^2)}\right)}{\frac{1}{\sqrt{2\pi}\sigma_Y}\exp\left(-\frac{1}{2}v^2\right)}\\\qquad\qquad\qquad\qquad=\frac{1}{\sqrt{2\pi(1-\rho^2)}\sigma_X}\exp\left(-\frac 12\left[\frac{u-\rho v}{\sqrt{1-\rho^2}}\right]^2\right)\tag 2$$ Indeed $$f_{X|Y}(x,y)=\frac{1}{\sqrt{2\pi(1-\rho^2)}\sigma_X}\exp\left(-\frac{1}{2}\left[\frac{x-(\mu_X+\rho\frac{\sigma_Y}{\sigma_X}(y-\mu_Y)}{\sigma_X\sqrt{1-\rho^2}}\right]^2\right)\tag 3$$$$f_{X|Y}(x,y)=\frac{1}{\sqrt{2\pi(1-\rho^2)}\sigma_X}\exp\left(-\frac{1}{2}\left[\frac{x-(\mu_X+\rho\frac{\sigma_X}{\sigma_Y}(y-\mu_Y)}{\sigma_X\sqrt{1-\rho^2}}\right]^2\right)\tag 3$$ as a result $$\mathbb{E}[X|Y]=\mu_X+\rho\frac{\sigma_Y}{\sigma_X}(y-\mu_Y)\tag 4$$$$\mathbb{E}[X|Y]=\mu_X+\rho\frac{\sigma_X}{\sigma_Y}(y-\mu_Y)\tag 4$$ and $$\text{Var}(X|Y)=\sigma_X^2(1-\rho^2)\tag 5$$

Another approach

$$f_{X|Y}(x,y)=\frac{f_{X,Y}(x,y)}{f_{Y}(y)}\tag 1$$ Set $$u=\frac{x-\mu_X}{\sigma_X}$$ and $$v=\frac{y-\mu_Y}{\sigma_Y}$$ we have $$f_{X|Y}(x,y)=\frac{\frac{1}{2\pi\sigma_X\sigma_Y\sqrt{1-\rho^2}}\exp\left(-\frac{u^2-2\rho uv+v^2}{2(1-\rho^2)}\right)}{\frac{1}{\sqrt{2\pi}\sigma_Y}\exp\left(-\frac{1}{2}v^2\right)}\\\qquad\qquad\qquad\qquad=\frac{1}{\sqrt{2\pi(1-\rho^2)}\sigma_X}\exp\left(-\frac 12\left[\frac{u-\rho v}{\sqrt{1-\rho^2}}\right]^2\right)\tag 2$$ Indeed $$f_{X|Y}(x,y)=\frac{1}{\sqrt{2\pi(1-\rho^2)}\sigma_X}\exp\left(-\frac{1}{2}\left[\frac{x-(\mu_X+\rho\frac{\sigma_Y}{\sigma_X}(y-\mu_Y)}{\sigma_X\sqrt{1-\rho^2}}\right]^2\right)\tag 3$$ as a result $$\mathbb{E}[X|Y]=\mu_X+\rho\frac{\sigma_Y}{\sigma_X}(y-\mu_Y)\tag 4$$ and $$\text{Var}(X|Y)=\sigma_X^2(1-\rho^2)\tag 5$$

Another approach

$$f_{X|Y}(x,y)=\frac{f_{X,Y}(x,y)}{f_{Y}(y)}\tag 1$$ Set $$u=\frac{x-\mu_X}{\sigma_X}$$ and $$v=\frac{y-\mu_Y}{\sigma_Y}$$ we have $$f_{X|Y}(x,y)=\frac{\frac{1}{2\pi\sigma_X\sigma_Y\sqrt{1-\rho^2}}\exp\left(-\frac{u^2-2\rho uv+v^2}{2(1-\rho^2)}\right)}{\frac{1}{\sqrt{2\pi}\sigma_Y}\exp\left(-\frac{1}{2}v^2\right)}\\\qquad\qquad\qquad\qquad=\frac{1}{\sqrt{2\pi(1-\rho^2)}\sigma_X}\exp\left(-\frac 12\left[\frac{u-\rho v}{\sqrt{1-\rho^2}}\right]^2\right)\tag 2$$ Indeed $$f_{X|Y}(x,y)=\frac{1}{\sqrt{2\pi(1-\rho^2)}\sigma_X}\exp\left(-\frac{1}{2}\left[\frac{x-(\mu_X+\rho\frac{\sigma_X}{\sigma_Y}(y-\mu_Y)}{\sigma_X\sqrt{1-\rho^2}}\right]^2\right)\tag 3$$ as a result $$\mathbb{E}[X|Y]=\mu_X+\rho\frac{\sigma_X}{\sigma_Y}(y-\mu_Y)\tag 4$$ and $$\text{Var}(X|Y)=\sigma_X^2(1-\rho^2)\tag 5$$

edited body
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user16651
user16651

Another approach

$$f_{X|Y}(x,y)=\frac{f_{X,Y}(x,y)}{f_{Y}(y)}\tag 1$$ Set $$u=\frac{x-\mu_X}{\sigma_X}$$ and $$v=\frac{y-\mu_Y}{\sigma_Y}$$ we have $$f_{X|Y}(x,y)=\frac{\frac{1}{2\pi\sigma_X\sigma_Y\sqrt{1-\rho^2}}\exp\left(-\frac{u^2-2\rho uv+v^2}{2(1-\rho^2)}\right)}{\frac{1}{\sqrt{2\pi}\sigma_Y}\exp\left(-\frac{1}{2}v^2\right)}\\\qquad\qquad\qquad\qquad=\frac{1}{\sqrt{2\pi(1-\rho^2)}\sigma_X}\exp\left(-\frac 12\left[\frac{u-\rho v}{\sqrt{1-\rho^2}}\right]^2\right)\tag 2$$ Indeed $$f_{X|Y}(x,y)=\frac{1}{\sqrt{2\pi(1-\rho^2)}\sigma_X}\exp\left(-\frac{1}{2}\left[\frac{x-(\mu_X+\rho\frac{\sigma_Y}{\sigma_X}(y-\mu_Y)}{\sigma_X\sqrt{1-\rho^2}}\right]^2\right)\tag 3$$ as a result $$\mathbb{E}[X|Y]=\mu_X+\rho\frac{\sigma_X}{\sigma_Y}(y-\mu_Y)\tag 4$$$$\mathbb{E}[X|Y]=\mu_X+\rho\frac{\sigma_Y}{\sigma_X}(y-\mu_Y)\tag 4$$ and $$\text{Var}(X|Y)=\sigma_X^2(1-\rho^2)\tag 5$$

Another approach

$$f_{X|Y}(x,y)=\frac{f_{X,Y}(x,y)}{f_{Y}(y)}\tag 1$$ Set $$u=\frac{x-\mu_X}{\sigma_X}$$ and $$v=\frac{y-\mu_Y}{\sigma_Y}$$ we have $$f_{X|Y}(x,y)=\frac{\frac{1}{2\pi\sigma_X\sigma_Y\sqrt{1-\rho^2}}\exp\left(-\frac{u^2-2\rho uv+v^2}{2(1-\rho^2)}\right)}{\frac{1}{\sqrt{2\pi}\sigma_Y}\exp\left(-\frac{1}{2}v^2\right)}\\\qquad\qquad\qquad\qquad=\frac{1}{\sqrt{2\pi(1-\rho^2)}\sigma_X}\exp\left(-\frac 12\left[\frac{u-\rho v}{\sqrt{1-\rho^2}}\right]^2\right)\tag 2$$ Indeed $$f_{X|Y}(x,y)=\frac{1}{\sqrt{2\pi(1-\rho^2)}\sigma_X}\exp\left(-\frac{1}{2}\left[\frac{x-(\mu_X+\rho\frac{\sigma_Y}{\sigma_X}(y-\mu_Y)}{\sigma_X\sqrt{1-\rho^2}}\right]^2\right)\tag 3$$ as a result $$\mathbb{E}[X|Y]=\mu_X+\rho\frac{\sigma_X}{\sigma_Y}(y-\mu_Y)\tag 4$$ and $$\text{Var}(X|Y)=\sigma_X^2(1-\rho^2)\tag 5$$

Another approach

$$f_{X|Y}(x,y)=\frac{f_{X,Y}(x,y)}{f_{Y}(y)}\tag 1$$ Set $$u=\frac{x-\mu_X}{\sigma_X}$$ and $$v=\frac{y-\mu_Y}{\sigma_Y}$$ we have $$f_{X|Y}(x,y)=\frac{\frac{1}{2\pi\sigma_X\sigma_Y\sqrt{1-\rho^2}}\exp\left(-\frac{u^2-2\rho uv+v^2}{2(1-\rho^2)}\right)}{\frac{1}{\sqrt{2\pi}\sigma_Y}\exp\left(-\frac{1}{2}v^2\right)}\\\qquad\qquad\qquad\qquad=\frac{1}{\sqrt{2\pi(1-\rho^2)}\sigma_X}\exp\left(-\frac 12\left[\frac{u-\rho v}{\sqrt{1-\rho^2}}\right]^2\right)\tag 2$$ Indeed $$f_{X|Y}(x,y)=\frac{1}{\sqrt{2\pi(1-\rho^2)}\sigma_X}\exp\left(-\frac{1}{2}\left[\frac{x-(\mu_X+\rho\frac{\sigma_Y}{\sigma_X}(y-\mu_Y)}{\sigma_X\sqrt{1-\rho^2}}\right]^2\right)\tag 3$$ as a result $$\mathbb{E}[X|Y]=\mu_X+\rho\frac{\sigma_Y}{\sigma_X}(y-\mu_Y)\tag 4$$ and $$\text{Var}(X|Y)=\sigma_X^2(1-\rho^2)\tag 5$$

edited body
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user16651
user16651

Another approach

$$f_{X|Y}(x,y)=\frac{f_{X,Y}(x,y)}{f_{X}(x)}\tag 1$$$$f_{X|Y}(x,y)=\frac{f_{X,Y}(x,y)}{f_{Y}(y)}\tag 1$$ Set $$u=\frac{x-\mu_X}{\sigma_X}$$ and $$v=\frac{y-\mu_Y}{\sigma_Y}$$ we have $$f_{X|Y}(x,y)=\frac{\frac{1}{2\pi\sigma_X\sigma_Y\sqrt{1-\rho^2}}\exp\left(-\frac{u^2-2\rho uv+v^2}{2(1-\rho^2)}\right)}{\frac{1}{\sqrt{2\pi}\sigma_X}\exp\left(-\frac{1}{2}u^2\right)}\\\qquad\qquad\qquad\qquad=\frac{1}{\sqrt{2\pi(1-\rho^2)}\sigma_Y}\exp\left(-\frac 12\left[\frac{v-\rho u}{\sqrt{1-\rho^2}}\right]^2\right)\tag 2$$$$f_{X|Y}(x,y)=\frac{\frac{1}{2\pi\sigma_X\sigma_Y\sqrt{1-\rho^2}}\exp\left(-\frac{u^2-2\rho uv+v^2}{2(1-\rho^2)}\right)}{\frac{1}{\sqrt{2\pi}\sigma_Y}\exp\left(-\frac{1}{2}v^2\right)}\\\qquad\qquad\qquad\qquad=\frac{1}{\sqrt{2\pi(1-\rho^2)}\sigma_X}\exp\left(-\frac 12\left[\frac{u-\rho v}{\sqrt{1-\rho^2}}\right]^2\right)\tag 2$$ Indeed $$f_{X|Y}(x,y)=\frac{1}{\sqrt{2\pi(1-\rho^2)}\sigma_Y}\exp\left(-\frac{1}{2}\left[\frac{y-(\mu_Y+\rho\frac{\mu_X}{\mu_Y}(x-\mu_X)}{\sigma_Y\sqrt{1-\rho^2}}\right]^2\right)\tag 3$$$$f_{X|Y}(x,y)=\frac{1}{\sqrt{2\pi(1-\rho^2)}\sigma_X}\exp\left(-\frac{1}{2}\left[\frac{x-(\mu_X+\rho\frac{\sigma_Y}{\sigma_X}(y-\mu_Y)}{\sigma_X\sqrt{1-\rho^2}}\right]^2\right)\tag 3$$ as a result $$\mathbb{E}[X|Y]=\mu_Y+\rho\frac{\mu_X}{\mu_Y}(x-\mu_X)\tag 4$$$$\mathbb{E}[X|Y]=\mu_X+\rho\frac{\sigma_X}{\sigma_Y}(y-\mu_Y)\tag 4$$ and $$\text{Var}(X|Y)=\sigma_Y^2(1-\rho^2)\tag 5$$$$\text{Var}(X|Y)=\sigma_X^2(1-\rho^2)\tag 5$$

Another approach

$$f_{X|Y}(x,y)=\frac{f_{X,Y}(x,y)}{f_{X}(x)}\tag 1$$ Set $$u=\frac{x-\mu_X}{\sigma_X}$$ and $$v=\frac{y-\mu_Y}{\sigma_Y}$$ we have $$f_{X|Y}(x,y)=\frac{\frac{1}{2\pi\sigma_X\sigma_Y\sqrt{1-\rho^2}}\exp\left(-\frac{u^2-2\rho uv+v^2}{2(1-\rho^2)}\right)}{\frac{1}{\sqrt{2\pi}\sigma_X}\exp\left(-\frac{1}{2}u^2\right)}\\\qquad\qquad\qquad\qquad=\frac{1}{\sqrt{2\pi(1-\rho^2)}\sigma_Y}\exp\left(-\frac 12\left[\frac{v-\rho u}{\sqrt{1-\rho^2}}\right]^2\right)\tag 2$$ Indeed $$f_{X|Y}(x,y)=\frac{1}{\sqrt{2\pi(1-\rho^2)}\sigma_Y}\exp\left(-\frac{1}{2}\left[\frac{y-(\mu_Y+\rho\frac{\mu_X}{\mu_Y}(x-\mu_X)}{\sigma_Y\sqrt{1-\rho^2}}\right]^2\right)\tag 3$$ as a result $$\mathbb{E}[X|Y]=\mu_Y+\rho\frac{\mu_X}{\mu_Y}(x-\mu_X)\tag 4$$ and $$\text{Var}(X|Y)=\sigma_Y^2(1-\rho^2)\tag 5$$

Another approach

$$f_{X|Y}(x,y)=\frac{f_{X,Y}(x,y)}{f_{Y}(y)}\tag 1$$ Set $$u=\frac{x-\mu_X}{\sigma_X}$$ and $$v=\frac{y-\mu_Y}{\sigma_Y}$$ we have $$f_{X|Y}(x,y)=\frac{\frac{1}{2\pi\sigma_X\sigma_Y\sqrt{1-\rho^2}}\exp\left(-\frac{u^2-2\rho uv+v^2}{2(1-\rho^2)}\right)}{\frac{1}{\sqrt{2\pi}\sigma_Y}\exp\left(-\frac{1}{2}v^2\right)}\\\qquad\qquad\qquad\qquad=\frac{1}{\sqrt{2\pi(1-\rho^2)}\sigma_X}\exp\left(-\frac 12\left[\frac{u-\rho v}{\sqrt{1-\rho^2}}\right]^2\right)\tag 2$$ Indeed $$f_{X|Y}(x,y)=\frac{1}{\sqrt{2\pi(1-\rho^2)}\sigma_X}\exp\left(-\frac{1}{2}\left[\frac{x-(\mu_X+\rho\frac{\sigma_Y}{\sigma_X}(y-\mu_Y)}{\sigma_X\sqrt{1-\rho^2}}\right]^2\right)\tag 3$$ as a result $$\mathbb{E}[X|Y]=\mu_X+\rho\frac{\sigma_X}{\sigma_Y}(y-\mu_Y)\tag 4$$ and $$\text{Var}(X|Y)=\sigma_X^2(1-\rho^2)\tag 5$$

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