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I want to come up with at least the expectation, and at best, the cdf, for a variable $Z$ that I think of as the result of a process and am not quite sure how to translate into equations.

Let $F(x) = x$, where $F(x)$ is the cdf defined over $(0,1)$ of random variable $X$ (uniform distribution).

Now let $G(y) = f1(F(y))$ where $f1$ in this case is just some function (I figure I don't need to write out the whole thing), with $Y\sim G(y)$, [EDIT: and $G$ is a valid cdf ($f1$ is a specific function that preserves cdf properties, I'm just using $f1$ for shorthand so I don't complicate the question with a long complicated function)].

Similarly, $H(w) = f2(F(w))$ and $W\sim H(w)$, [EDIT: and $H$ is a valid cdf ($f2$ is a specific function that preserves cdf properties, I'm just using $f2$ for shorthand so I don't complicate the question with a long complicated function)]

Now, I want to define a random variable $Z$ such that $Z$ is a weighted average of either a draw from $G$ (with probability $a$) or a draw from $H$ that is strictly greater than whatever the draw from $G$ was (with probability $1-a$).

So, in other words, I want to write out the distribution of $Z$ where $Z$ looks something like (I know this is not quite correct form) $Z= a Y + (1-a)(W|W>Y)$

Can anyone help me out?

Thanks so much!

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    @DilipSarwate, Actually, I am a *her*, and thank you for making me be more clear on the issue.2012-02-15

2 Answers 2

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Call $f_Y$ the probability density function of $Y$, $f_W$ the probability density function of $W$, and $F_W$ the cumulative density function of $W$. Then the probability density function $f_Z$ of $Z$ is $ f_Z(z)=af_Y(z)+(1-a)f_W(z)\int_{-\infty}^z\frac{f_Y(y)\mathrm dy}{1-F_W(y)}. $ Equivalently, the cumulative density function $F_Z$ of $Z$ is $ F_Z(z)=F_Y(z)-(1-a)(1-F_W(z))\int_{-\infty}^z\frac{f_Y(y)\mathrm dy}{1-F_W(y)}. $

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    Because when one integrates the first displayed equation on $(-\infty,z)$ (something you surely did by now), one obtains the second displayed equation with no factor $a$.2012-02-25
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May apologies to Jand for referring to her as a male. Comments are restricted in length and I failed to write them in a gender-neutral way.

The question raised still seems very confusing to me. There is a totally useless random variable $X$ whose sole purpose seems to be to introduce $F$ which is the identity function from $(0,1)$ to $(0,1)$ since $F(x)=x$ for $x \in (0,1)$. Then, $G(y) = f1(F(y)) = f1(y)$ as far as I can tell, or at least for $y \in (0,1)$, where we are assured that $G$ is a CDF. Similarly for $H$. Now $Y$ and $W$ are defined as random variables with cumulative distribution functions $G$ and $H$ respectively. Their joint distribution is not specified but let us assume that they are independent. From these is constructed a new random variable $Z$, a function of $(Y,W)$ about which it is said that

$Z$ is a weighted average of either a draw from $G$ (with probability $a$) or a draw from $H$ that is strictly greater than whatever the draw from $G$ was (with probability $1−a$).

My interpretation is that there is yet another random variable $V$ which is Bernoulli$(a)$ and presumably independent of $Y$ and $W$, and we have

$Z = Z(Y,W,V) = \begin{cases} Y, &\text{if}~ V = 1,\\ W, &\text{if}~ V = 0, ~\text{and}~W > Y,\\ ?? &\text{if}~ V = 0, ~\text{and}~W \leq Y. \end{cases}$ In words, what happens if our biased coin came down Tails (this happens with probability $(1-a)$) meaning that we are suppose to use $W$, the draw from $H$, if possible, but using $W$ is not permitted because $W$ is not strictly larger than the draw $Y$ from $G$? Are $W$ and $Y$ drawn again and again till we end up with a $W > Y$?

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    I am trying to explain this as clearly as I can but I lack the appropriate training and terminology. Thank you for helping me figure this out.2012-02-15