Truncated gaussian

Question

Let $X$ a random variable with density $f_X(x)\propto \frac{c}{\sqrt{2\pi}}e^{-\frac{x^2}{2}},x\in [-1,1]$, and let $Y|X=x \sim N(x,1)$. Find the value of $c$.

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I'm really struggling the notation of proportionality: I don't understand how to approach the problem. Usually, when I have a random vector and I have to find the value of a constant I apply the standardisation condition $ 1=\int_{\mathbb{R}^2}f_{XY}(x,y)dxdy$, but in this case I get confused by the symbol $\propto$. Clearly I have to apply the definition $f_{Y|X}(x|y):=\frac{f_{XY}(x.y)}{f_X(x)}\Rightarrow f_{XY}(x,y)=f_X(x)f_{Y|X}(x|y)$. Can you help me? Thanks in advance.

Answer 1

I am confuse with your text

Let $X$ a random variable with density $f_X(x)\propto \frac{c}{\sqrt{2\pi}}e^{-\frac{x^2}{2}},x\in [-1,1]$, and let $Y|X=x \sim N(x,1)$. Find the value of $c$.

If the marginal X density is truncated in $x \in[-1;1]$ the constant can be easily found by integrating the density all over its domain

$$c\int_{-1}^{1}\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx=1$$

$$c[\Phi(1)-\Phi(-1)]=1$$

$$c=\frac{1}{[\Phi(1)-\Phi(-1)]}=\frac{1}{[2\Phi(1)-1]}\approx 1,46479477349154$$

I do not understand why the text gives you also the conditional $(Y|X)$ density...perhaps is for a further question?

Answer 2

@tommik Always for completeness:

2. $\mathbb{E}(X)=1, Var(X)=1\Rightarrow X\sim N(0,1)$;

3. $f_Y(y)=\frac{1}{\sqrt{2\pi}}\Rightarrow Y\sim N(x,1)$;

4. $Cov(X,Y)=\mathbb{E}[XY]-\mathbb{E}[X]\mathbb{E}[Y]=\mathbb{E}[XY]=\int_{\mathbb{R}^2}xyf_{XY}(x,y)dxdy=\int_{\mathbb{R}}xf_X(x)dx[\int_{\mathbb{R}}yf_{Y|X}(x|y)dy]dx=\mathbb{E}[x\mathbb{E}[Y|X]]=\mathbb{E}[XX]=\mathbb{E}[X^2]=1$

Answer 3

$X$ is a truncated standard gaussian in $[-1;1]$ so its mean is 0. With simple calculations (integration by parts) you get $E[X^2]\approx 0,29=cov (X,Y)$

Y density is not easy to be calculated