Could someone please point me to a source or suggest ways in which we can obtain the Distribution, Density Functions, Expected Value, etc. of a Uniform Distribution whose parameters are distributed Normally.


$$W \sim U[X,Y]$$

$$X \sim N[\mu_{X},\sigma^2_{X}]$$

$$Y \sim N[\mu_{Y},\sigma^2_{Y}]$$

$X,Y$ can be assumed to be independent if it simplifies matters. But as Dilip points out, when $X>Y$ there is confusion since Uniform Distrbution is undefined when $X>Y$. Is there some sort of right truncated and left trucated extensions of the normal distribution that we can use for $X,Y$. Please suggest how we can resolve this issue as well.

To Determine,

$$f_{W}(w), F_{W}(w), E(W)$$

Please note, I have searched both on this forum and elsewhere without finding much relevant material. My understanding of these concepts is fairly basic. If this is a redundant question, please let me know and I am happy to delete this to reduce the noise in the forum.

Related General Question

Starting with the above special case, it quickly becomes apparent there are many combinations possible. Hence was wondering if there were general techniques to derive the density, distribution function, expected value, higher moments, conditional expectations etc. of compound distributions and some source where certain combinations and results therein were given with detailed steps and complete proofs: https://math.stackexchange.com/questions/1614212/compound-distributions-basic-techniques-and-key-general-results-from-first-p

  • $\begingroup$ What happens if $X > Y$? Are $X$ and $Y$ independent? $\endgroup$ – Dilip Sarwate Jan 13 '16 at 7:49
  • $\begingroup$ Can you assume independence of $X$ and $Y$? $\endgroup$ – sinbadh Jan 13 '16 at 7:50
  • $\begingroup$ Independence simplifies matters I suppose ... I will add this to the question. $\endgroup$ – texmex Jan 13 '16 at 7:51
  • $\begingroup$ @probablyme ... True Uniform Distrbution is undefined when $X > Y$ ... How could we handle this ? Is there some sort of right truncated and left trucated extensions of the normal distribution that we can use for $X,Y$ $\endgroup$ – texmex Jan 13 '16 at 7:57

Let $w$ fixed. Now, by Total Probability:

$\begin{eqnarray*} &&P(W\le w)\\ &=&P(W\le w|X<Y)P(X<Y)+P(W\le w|X>Y)P(X>Y)\\ &=&P(W\le w|X\mbox{ and }Y>w)P(X\mbox{ and }Y>w)P(X<Y)\\ &&+P(W\le w|X<w<Y)P(X<w<Y)P(X<Y)\\ &&+P(W\le w|Y<w<X)P(Y<w<X)P(X<Y)\\ &&+P(W\le w|X\mbox{ and }Y<w)P(X\mbox{ and }Y<w)P(X<Y) \end{eqnarray*}$

Now, since $X$ and $Y$ satisfie $X\le Y$ (by conditioning), then third conitional is 0. Also, the first one is 0, cause can't happen $W\le w$ given $X,Y>w$. Finally, in fourth one $P(W\le w|X\mbox{ and }Y<w)=1$ cause $X\le W\le Y<w$. Then,

$$P(W\le w)=[P(W\le w|X<w<Y)P(X<w<Y)+P(X\mbox{ and }Y<w)]P(X<Y)$$

which can be easly calculated (remembering that, implicity, we are supposing $X<Y$ given. Thus, for example, $P(X\mbox{ and }Y<w)$ indeed is $P(X\mbox{ and }Y<w|X<Y)$)

  • $\begingroup$ Thanks for your answer and I suppose step two to step three is from the law of total probability. Also, the area of compound distributions seems to be less studied and hence I have posted a more general question on this (Linked into question now). Could you please offer any relevant advice on that? $\endgroup$ – texmex Jan 16 '16 at 13:07
  • $\begingroup$ Since X and Y are normally distributed random variables; is it valid to make the assumption that X<Y? Or under what conditions or criteria can we make this assumption? Once this assumption is made, we have the cumulative distribution function, can we just differentiate to get the density function and follow standard methods for expectation and second moment etc? What can we day about the distribution of W? It is not normal or uniform or any common well known distribution? Please let me know if any of this is not clear. $\endgroup$ – texmex Jul 27 '16 at 5:55

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