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Suppose I have one normal distribution $W \sim N(\mu_{w},\sigma_{w}^2)$ with a known cuttoff point (percentile) on this distribution called $c$. The first part of $W \in [-\infty,c[$ needs to be convoluted with another normal distribution $X\sim N(\mu_{x},\sigma_{x}^2)$ and the second part of $W \in [c,\infty]$ needs to be convoluted with a Normal distribution $Y\sim N(\mu_{y},\sigma_{y}^2)$. What are the properties of the resulting distribution $Z$?

Is this somehow similar to the problem of the sum of a normal distribution and a truncated distribution described in http://www.jstor.org/stable/1266101?seq=2

Or is this equal to the sum of truncated normal distributions?

More specifically in my application $Z$ represents an arrival time distribution and $W$ represents a departure time distribution and $c$ the end of a timeslot. The arrival time for departures smaller than $c$ is the sum of $W$ and $X$ and the arrival time for departures bigger than $c$ is the sum of $W$ and $Y$.

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I think this is a very nice problem. If I may change notation slightly ...

The Problem

Let $\quad W \sim N(\mu_0, \sigma_0^2), \quad X_1 \sim N(\mu_1, \sigma_1^2), \quad X_2 \sim N(\mu_2, \sigma_2^2)$

denote independent random variables, and let $c$ denote a constant.

Find the pdf of $Z$, where:

$$ Z = \begin{cases}W + X_1 & \text{if } W \leq c \\ W + X_2 & \text{if } W > c \end{cases}$$

Solution

To solve this, we need to solve 2 problems.

  1. Find $h_1(z)$: the pdf of $(W + X_1) \, \big| \, (W \leq c) \quad $ (i.e. truncated-above Normal + Normal)

  2. Find $h_2(z)$: the pdf of $(W + X_2) \, \big| \, (W > c) \quad $ (i.e. truncated-below Normal + Normal)

Then the pdf of $Z$, say $h(z)$, is the component mix:

$$h(z) \, = \, P(W \leq c) * h_1(z) \quad + \quad P(W>c) * h_2(z)$$


Solution: Part 1

$\rightarrow$ The pdf of the sum of a truncated-above Normal and a Normal

If $W$ is truncated ABOVE at $c$, ... then the joint pdf of $(W \big|(W \leq c),X_1)$, say $f_1(w,x_1;c)$, is, by independence, simply the product of the respective individual pdf's ... that is, $f_1(w,x_1;c) = \frac{f_w(w)}{P(W<c)} * f_{x_1}(x_1)$:

enter image description here

Next, transform $(W,X_1) \rightarrow (Z=W+X_1, V=X_1)$. Here is the joint pdf of $(Z, V)$, say $g_1(z,v)$:

enter image description here

where:

  • I am using the Transform function in the mathStatica package for Mathematica to do the nitty-gritties.

  • Note that the transformation equation $(Z=W+X_1, V=X_1)$ induces dependency between $Z$ and $V$. In particular, since $Z=V+W$ and $W < c$, it follows that $Z < V + c$. This important constraint is entered using the Boole[ blah ] statement above.

  • Erf[.] denotes the error function

We seek the marginal pdf of $Z = W + X_1$, say $h_1(z)$, which is:

enter image description here

... defined on the real line. This concludes Part 1.


Solution: Part 2

$\rightarrow$ The pdf of the sum of a truncated-below Normal and a Normal

If $W$ is truncated BELOW at $c$, ... then the joint pdf of $(W \big|(W > c),X_2)$, say $f_2(w,x_2;c)$, is, by independence, simply the product of the respective individual pdf's ... that is, $f_2(w,x_2;c) = \frac{f_w(w)}{P(W>c)} * f_{x_2}(x_2)$:

enter image description here

Next, transform $(W,X_2) \rightarrow (Z=W+X_2, V=X_2)$. Here is the joint pdf of $(Z, V)$, say $g_2(z,v)$:

enter image description here

  • Note that the transformation equation $(Z=W+X_2, V=X_2)$ induces dependency between $Z$ and $V$. In particular, since $Z=V+W$ and $W > c$, it follows that $Z > V + c$. This important constraint is entered using the Boole[ blah ] statement above.

We seek the marginal pdf of $Z = W + X_2$, say $h_2(z)$, which is:

enter image description here

...defined on the real line. This concludes Part 2.


The Component Mix

All the necessary pieces to the puzzle are now in place. To make this explicit, if $W \sim N(\mu_0, \sigma_0^2)$ with pdf $f(w)$:

enter image description here

... then $P(W<c)$ is:

enter image description here

Recall that the pdf of $Z$ is:

$$h(z) \, = \, P(W \leq c) * h_1(z) \quad + \quad P(W>c) * h_2(z)$$

... which is explicitly:

enter image description here

where $Z$ is defined on the real line. All done.


Monte Carlo check

It is always a good idea to check symbolic work using alternative methods. Here is a quick Monte Carlo check when:

$\text{params}=\left\{\mu _0\to 16,\mu _1\to 3,\mu _2\to 2,\sigma _0\to 6,\sigma _1\to 0.1,\sigma _2\to 2,c\to 12\right\}$

The following plot compares:

  • a Monte Carlo simulation of the pdf of $Z$ (squiggly BLUE curve) to the
  • theoretical solution derived above (dashed RED curve)

enter image description here

Looks fine :) Different parameter choices can, of course, yield very different shaped outcomes.


Mean of $Z$

Paulius Šarka asks: "Does the mean of Z have an analytical form"

Yes - it is easiest to derive this from:

$$ Z = \begin{cases}W + X_1 & \text{if } W \leq c \\ W + X_2 & \text{if } W >c\end{cases}$$

... it follows that:

$$E[Z] = P(W \leq c) \big(E[W \big | W \leq c] + \mu_1 \big) \quad + \quad P(W>c)\big(E[W \big | W > c] + \mu_2 \big)$$

which yields the closed-form solution:

$$E[Z] \quad = \quad \mu_0 \, + \, P(W \leq c) \mu_1 \, + \, P(W > c) \mu_2$$

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    $\begingroup$ Nice! Does the mean of $Z$ have an analytical form (i.e. can Mathematica integrate that expression)? $\endgroup$ – psarka Oct 31 '14 at 16:36
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    $\begingroup$ Hi @PauliusŠarka Yes - the mean has a 'nice' closed-form -- see addendum above -- best derived from the defn of $Z$. $\endgroup$ – wolfies Oct 31 '14 at 17:59
  • $\begingroup$ @wolfies Concerning the analytical form for the mean of $Z$, why is $E[W|W \leq c]$ equal to $E[W]=\mu_{0}$? $\endgroup$ – cevertje400 Nov 19 '14 at 14:05
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    $\begingroup$ $E[W \big | W \leq c]$ is not equal to $E[W]$. The correct derivation is: $$blah = P(W \leq c) \big(E[W \big | W \leq c] \big) \quad + \quad P(W>c)\big(E[W \big | W > c] \big) = E[W] = \mu_0$$ IF this is not intuitive, more formally let $W$ have pdf $\phi(w)$. Then $$\phi(w \big| W \leq c) =\frac{\phi(w)}{P(W \leq c)} \text{and} E[W \big | W \leq c] = \int_{-\infty}^c w \frac{\phi(w)}{P(W \leq c)}dw$$ Similarly, $E[W \big | W >c] = \int_{c}^\infty w \frac{\phi(w)}{P(W > c)}dw$. So, $$blah = \int_{-\infty}^c w \phi(w) dw + \int_{c}^\infty w \phi(w)dw = E[W]$$ $\endgroup$ – wolfies Nov 19 '14 at 17:49

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