Calculation of an "unconstrained" normal distribution (starting from a censored one)

Question

Assume that two r.v. $W$ and $Y|W=w$ with

(1) $W \sim \text{N}(\mu_w,\sigma_w^2)$ (iid)

(2) $Y|W=w \sim \text{N}(w,\sigma_y^2)$ (iid)

Further we only observe $Y$ if $Y$ is less then $W$, i.e.,

(3) $Y|Y\le W$

Goal: Find the pdf of the censored observations, i.e., of $Y|Y\le W$ and from that deduce the uncensored pdf and the first two moments (so i.m.h.o. we have to find$f_Y(y)$). The first two moments of this uncensored pdf are supposed to depend upon $E(Y|Y\le W)$ and $Var(Y|Y\le W)$.

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By definition of conditional pdf we have that:

(4) $f_{Y|W}(y|W = w)= \frac{f_{Y,W}(y,w)}{f_W(w)}$

Next, the definition of a truncated density gives for a abitrary value of $W$:

(5) $ f_{Y|Y\le W}(y|y\le w) = \frac{f_Y(y)}{P(Y\le W)}$

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I would simply rewrite (4) to

$f_{Y|W}(y|W = w)f_W(w) = f_{Y,W}(y,w)$

then integration over $f_{Y,W}(y,w)$ w.r.t $w$ should yield $f_Y(y)$, i.e.,

(a) $\int_{-\infty}^{\infty} f_{Y,W}(y,w) dw = \int_{-\infty}^{\infty} f_Y(y|W = w)f_W(w) dw = f_Y(y)$

Plugin in $f_Y(y)$ into (5), ($P(Y\le W)$ will also be given by $f_Y(y)$) I will se how the moments of $f_{Y|Y\le W}(y|y\le w)$ will look and how the moments of $f_Y(y)$ depend upon them.

So (a) will look like

$f_Y(y) = \int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi\sigma^2_y}}\text{exp}\big(-\frac{(y-w)^2}{2\sigma_y^2}\big)\frac{1}{\sqrt{2\pi\sigma^2_w}}\text{exp}\big(-\frac{(w-\mu_w)^2}{2\sigma_w^2}\big)dw$

Except for the $w$ in the first $\text{exp}$, this looks very easy but since there is a $w$ im a little bit stuck how to solve this...

Answer 1

Ok. Let's do this, for CV's shake.

First compact by setting $C=\frac{1}{\sqrt{2\pi\sigma^2_y}}\frac{1}{\sqrt{2\pi\sigma^2_w}} = \frac{1}{2\pi\sigma_y\sigma_w}$, so

$$f_Y(y) =C \int_{-\infty}^{\infty}\exp\left\{-\frac{(y-w)^2}{2\sigma_y^2}\right\}\exp\left\{-\frac{(w-\mu_w)^2}{2\sigma_w^2}\right\}dw$$

We have $$exp\left\{-\frac{(y-w)^2}{2\sigma_y^2}\right\}\exp\left\{-\frac{(w-\mu_w)^2}{2\sigma_w^2}\right\} = \exp\left\{-\frac{y^2-2yw+w^2}{2\sigma_y^2}\right\}\exp\left\{-\frac{w^2-2w\mu_w+\mu_w^2}{2\sigma_w^2}\right\} =\exp\left\{-\frac{y^2}{2\sigma_y^2}-\frac{\mu_w^2}{2\sigma_w^2}\right\} \exp\left\{-\frac{w^2}{2\sigma_y^2}-\frac{w^2}{2\sigma_w^2}\right\}\exp\left\{\frac{2yw}{2\sigma_y^2}+\frac{2w\mu_w}{2\sigma_w^2}\right\}$$

Setting $s^2\equiv \sigma_y^2+\sigma_w^2$ we arrive at

$$=\exp\left\{-\frac{y^2}{2\sigma_y^2}-\frac{\mu_w^2}{2\sigma_w^2}\right\} \exp\left\{-\frac{s^2}{2\sigma_y^2\sigma_w^2}w^2\right\}\exp\left\{\frac{\sigma_w^2y+\sigma_y^2\mu_w}{\sigma_y^2\sigma_w^2}w\right\}$$

Include the first $\exp$ in the constant, $C^*=C \exp\left\{-\frac{y^2}{2\sigma_y^2}-\frac{\mu_w^2}{2\sigma_w^2}\right\}$. Set $$\beta\equiv \frac{s^2}{2\sigma_y^2\sigma_w^2},\qquad \alpha\equiv \frac{\sigma_w^2y+\sigma_y^2\mu_w}{\sigma_y^2\sigma_w^2}$$ to obtain

$$f_Y(y) =C^* \int_{-\infty}^{\infty}e^{-\beta w^2+\alpha w}dw=C^*\left[ \int_{-\infty}^{0}e^{-\beta w^2+\alpha w}dw + \int_{0}^{\infty}e^{-\beta w^2+\alpha w}dw\right]$$

$$=C^* \int_{0}^{\infty}e^{-\beta w^2}\left[e^{-\alpha w}+e^{\alpha w}\right]dw =2C^* \int_{0}^{\infty}e^{-\beta w^2}\operatorname{cosh}(\alpha w)dw$$

where $\operatorname{cosh}$ is the hyperbolic cosine.

Using a formula provided in Gradshteyn & Ryzhik (2007), "Table of Integrals, Series and Products", 7th ed., p. 384, eq. 3.546(2) we have

$$f_Y(y)=2C^*\frac 12 \sqrt {\frac {\pi}{\beta}} \exp\left\{\frac {\alpha^2}{4\beta}\right\}$$

Now $$\frac {\alpha^2}{4\beta} = \frac {\left(\frac{\sigma_w^2y+\sigma_y^2\mu_w}{\sigma_y^2\sigma_w^2}\right)^2}{4\frac{s^2}{2\sigma_y^2\sigma_w^2}} = \frac {(\sigma_w^2y+\sigma_y^2\mu_w)^2}{2\sigma_y^2\sigma_w^2s^2}$$

and bringing back in $C^*$ (and $\beta$) in all its glory we have

$$f_Y(y)=\frac{1}{2\pi\sigma_y\sigma_w}\exp\left\{-\frac{y^2}{2\sigma_y^2}-\frac{2\mu_w^2}{\sigma_w^2}\right\}\sqrt{\pi} \left(\sqrt {\frac{s^2}{2\sigma_y^2\sigma_w^2}}\right)^{-1} \exp\left\{\frac {(\sigma_w^2y+\sigma_y^2\mu_w)^2}{2\sigma_y^2\sigma_w^2s^2}\right\} $$

The constant terms simplify to

$$\frac{1}{2\pi\sigma_y\sigma_w}\sqrt{\pi} \left(\sqrt {\frac{s^2}{2\sigma_y^2\sigma_w^2}}\right)^{-1} = \frac{1}{s\sqrt{2\pi}} $$

and, the exponentials end up in the normal exponential. So in the end

$$f_Y(y) = \frac{1}{s\sqrt{2\pi}}\exp\left\{-\frac{(y-\mu_w)^2}{2s^2}\right\}= N(\mu_w, s^2),\qquad s^2\equiv \sigma_y^2+\sigma_w^2$$

Answer 2

$Y = W + Z$, where $Z$ is normal with mean 0 and variance $\sigma_z^2$ and is independent of $W$. (Note that I am using $\sigma_z^2$ where the OP used $\sigma_y^2$, which I reserve for the marginal variance of $Y$.) Then the unconstrained joint distribution of $(W,Y)$ is bivariate normal with $\mu_y = \mu_w$, $\sigma_y^2 = \sigma_w^2 + \sigma_z^2$, and $\sigma_{wy} = \sigma_w^2$.

Letting $\phi$ denote the standard normal pdf, integrating over the halfplane $Z<0$ gets the following marginal moments of $Y\,|\,(Y<W)$:

Mean $= \mu_y = 2 \int_{-\infty}^\infty \int_{-\infty}^0 (w \sigma_w + \mu_w + z \sigma_z)\,\phi(z)\mathrm{d}z\,\phi(w)\mathrm{d}w = \mu_w - \sigma_z \sqrt{2/\pi}$.

Variance $=\sigma_y^2 = 2 \int_{-\infty}^\infty \int_{-\infty}^0 (w \sigma_w + \mu_w + z \sigma_z - \mu_y)^2\,\phi(z)\mathrm{d}z\,\phi(w)\mathrm{d}w = \sigma_w^2 + \sigma_z^2 (1-2/\pi)$.

Third central moment $=2 \int_{-\infty}^\infty \int_{-\infty}^0 (w \sigma_w + \mu_w + z \sigma_z - \mu_y)^3\,\phi(z)\mathrm{d}z\,\phi(w)\mathrm{d}w = \sqrt{2}(\pi - 4) \sigma_z^3 / \pi^{3/2}$.

Those can be solved in reverse order to get $\sigma_z^2$, then $\sigma_w^2$, then $\mu_w$, which are necessary and sufficient to specify the unconstrained joint distribution.