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I have a function $Y=\displaystyle \sum_{i=1}^n X_i \Omega_i$, which represents the Isotope picked up after surfaces are touched. Here $X_i\sim N (\mu_i,\sigma_i)$. and $\Omega_i$ are constants of surface isotope concentration.

How can I calculate the conditional variance of the expectation of $Y$ given a particular $x_{i^*}$: ie $V_{x_i^*}(E[Y|X_i=x_{i^*}])$?. Can this be done analytically?

The data set produces for example: enter image description here

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Unless I'm misunderstanding the question, this is straightforward: Expectation is linear, so the expectation of all terms except for the $i^*$-th just becomes a constant that doesn't affect the variance, so this is just $V_{x_i^*}(E[X_i\Omega_i|X_i=x_{i^*}])$, which is $\Omega_i^2$ times the variance of $X_i$.

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    Could you elaborate on how you found that?2012-08-09
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    @user1134241: which part?2012-08-09
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    The analytical finding of the variance if possible.I'm attempting to calculate sensitivity indices $S_i$. If $X_i$ was drawn from data rather than a known continuous distribution, would this alter the outcome?2012-08-09
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    @user1134241: I still don't feel sure I understand your question correctly. There's not much analytical in my answer; I just applied the linearity of expectation and the translational invariance of the variance. I don't see how this could have anything to do with how you obtain the distribution for $X_i$. But I may just not be following what you do. Perhaps you could say more specifically what part of the answer you don't understand. The linearity? The translational invariance? The fact that the variance of $X\Omega$ is $\Omega^2$ times the variance of $X$?2012-08-09
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    The latter I think is probably it. Thanks2012-08-09
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    @user1134241: OK. From what you wrote I take it that $\Omega_i$ are fixed constants. For a random variable $X$, the variance is $V(X)=\langle(X-\langle X\rangle)^2\rangle$, so if $\Omega$ is a constant, then for $\Omega X$ this is $$V(\Omega X)=\langle(\Omega X-\langle \Omega X\rangle)^2\rangle=\langle(\Omega X-\Omega\langle X\rangle)^2\rangle=\langle(\Omega (X-\langle X\rangle))^2\rangle=\langle\Omega^2(X-\langle X\rangle)^2\rangle=\Omega^2\langle(X-\langle X\rangle)^2\rangle=\Omega^2V(X)\;.$$ (Here $\langle\cdot\rangle$ denotes taking the expectation value.)2012-08-09
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    Thank you. Is that not showing the unconditional variance however? If I refer to the scatter graph above right, the `x axis` can be split up into vertical slices and the expectation of values taken within each slice. Then if I take the variance of those expectations I should come up with the conditional variance of expectation when $X=x_i^*$, no?2012-08-09
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    @user1134241: I think you're going to have to say more about what that diagram shows for me to understand that; I don't see how the diagram relates to your formulas -- that was part of why I said I'm not sure I understand the question correctly. The calculation above holds for conditional and unconditional variance alike; whenever you multiply a random variable by a constant, the variance is multiplied by the square of the constant. The calculation wouldn't change if you add conditionalizations everywhere.2012-08-09
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    I have a vector if input parameters $X=X(n,pt)$ where n is the number of times a person touches a surface, $pt$ is the transfer efficiency of isotope from surface to hand. The right hand scatter plot is $Y$ vs $n$ for a sample of 1000 people. (ie the final amount of isotope on a person's hand after touching $n$ surfaces.) I'd like to find out which of these 2 parameters is causing the most variation in $Y$ and quantify it by sensitivity analysis.2012-08-12
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    @user1134241: I'm still having trouble relating that to what you wrote in the question. Let me rephrase what I've understood so far to see whether I got it right: You have $1000$ people. A person touches $n$ surfaces, where $n$ differs from person to person, and picks up isotope $i$ on surface $i$, with $1\le i\le n$. Somehow $X_i\Omega_i$ represents the amount of isotope $i$ picked up -- you haven't said anything about why this is split up that way, instead of just $X_i\sim N (\Omega_i\mu_i,\Omega_i^2\sigma_i)$ and $Y=\sum_iX_i$, but perhaps that's not important.2012-08-12
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    @user1134241: What I don't understand is where $pt$ comes into this. When you say you want to find out which of these $2$ parameters is causing the most variation in $Y$, do you mean the two parameters $n$ and $pt$? If so, don't you have to say something about how $pt$ enters into $Y$? It's turned up for the first time in that comment, without any apparent relationship to the rest of the question.2012-08-12
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    I apologise I was trying to use some standard notation that seems to crop up a lot in books. It seems that it's confused us both. You are 100% correct in your penultimate comment up until $1\leq i\n$. Isoptope on surface $i$ is $\Omega_i$. Each time surface $i$ is touched will allow a percentage of $\Omega$ to be transfer to the hand. This fraction is $pt$ (ie probability of transfer or transfer efficiency $0 \leq pt \leq 1 $). So from one surface contact $Y$ isotope will be picked up. Therefore from n surface contacts $Y=\sum^n X_i \Omega_i$. Does that make more sense?2012-08-12
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    @user1134241: A bit more. So when you say "a percentage of $\Omega$ ... This fraction is $pt$", are you saying that these were just two names that you used for the same quantity? If so, I don't understand why $pt$ was one parameter and $\Omega_i$ were $n$ parameters. This might be related to your use of $\Omega$ without a subscript, which I also don't understand, since $\Omega_i$ had only been introduced with subscripts, pertaining to individual surfaces.2012-08-12
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    $pt_i$ is the fraction of isotope $\Omega_i$ picked up each time the surface is touched. Eg. let's say surface 1 had $\Omega_1=100$ units of isotope on it and during one contact $pt_1$ was 2% so $y_1=0.02*100$. So $pt_i$ varies each time a surface is touched because you don't always pick up the same fraction each time. hence the subscript $pt_i$. This then happens for $n$ surfaces.2012-08-12
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    @user1134241: I'm afraid I'm understanding less and not more. If you don't mind me saying so, I think you should make an effort to state things with more precision. I don't understand what you meant by "I'd like to find out which of these $2$ parameters is causing the most variation in $Y$" -- it now seems you have not $2$ but $n+1$ parameters per person? (one parameter being $n$, the other $n$ parameters being the percentages $pt_i$) I also still don't understand how exactly $p_i$, $X_i$ and $\Omega_i$ are related. If $Y=\sum_iX_i\Omega_i$, then how does $pt_i$ enter into all this?2012-08-12
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    I see that you wrote $X=X(n,pt)$ at some point -- I don't know how I should relate that to the $X_i$, since you never mentioned $X$ without a subscript anywhere else, and if I add a subscript, $X_i=X_i(n,pt_i)$, I don't know what sense to make of the $n$ in there.2012-08-12
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    Ok let me clear the slate. I observe a person going around touching surfaces ($i$) in a room. The total number of surfaces they touch is $n$. Each surface is impregnated with an amount of isotope $\Omega_i$. They don't take up all the isotope each time they touch the surface, but a fraction $pt_i\times \Omega_i$. So at the end of the observation they have $\displaystyle \sum^n_ {i=1} pt_i\Omega_i$ isotope on their hands. I called this value Y. I thought I would avoid details and just write $X=X(n,pt)$, though I see this was misleading.2012-08-12
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    @user1134241: Ah -- so $X_i$ and $pt_i$ were two names for the same quantity? And $pt_i$ has normal distribution, as given for $X_i$ in the question?2012-08-12
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    @user1134241: If I now understand the setup correctly, then I still don't understand what you meant by "I'd like to find out which of these $2$ parameters is causing the most variation in $Y$". It now seems that you meant the $n+1$ parameters $n$ and $pt_i$; but since how many of the $pt_i$ there are depends on $n$, how do you define a separate causing of variation by $n$ and by the $pt_i$?2012-08-12
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    The variation in Y caused by $pt$ is given by the conditional variance of the expectation of $pt$ divided by the unconditional variance of Y: $\displaystyle \dfrac{V_{pt^*}(E[Y|pt^*])}{V(Y)}$. Where $pt^*$ is the fixed value of $pt$.2012-08-12
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    @user1134241: I think I'll be giving up on this soon, it's too time-consuming and frustrating, and I don't feel you're making a concerted effort to be precise. You write "the expectation of $pt$", but the formula contains an expectation of $Y$ under a condition on $pt^*$. And you're again using $pt$ without subscripts, despite my question being specifically about the problem of there being a varying number of $pt_i$, which can't be addressed by collectively referring to all of them as "$pt$".2012-08-12
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    I apologise, my own mistake in subscripting $pt$ has made this unnecessarily complicated. Each time a surface is touched a value of $pt$ is drawn from a distribution. Then there are only 2 parameters in the model $pt$ and $n$ oder?2012-08-12
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    @user1134241: So both $X_i$ and $pt_i$ were actually intended to be the same scalar quantity $pt$, so in fact $Y=\sum_ipt\Omega_i=pt\sum_i\Omega_i$? For each person, we have $n$ and $pt$. Then your formulation in the question "given a particular $x_{i^*}$" no longer makes any sense. What I still don't understand is, it seems that *which* $n$ surfaces the person touched, and hence the values of the $\Omega_i$ in the sum, is also relevant? So it's still not really the case that we have only two parameters. But now it's at least possible to focus on those two parameters, if that's what you want.2012-08-12
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    Stellen Sie sich mal vor Sie befinden sich in einem Zimmer wo nur einen Tisch, ein Bett, einen Regal und ein Seseel drin stehen. Dieses mal fassen Sie nur drei von den vier Möbel an. Ihre Hände nehmen durch jeden Kontakt eine bestimmte Menge Isotop an. Was aber nicht immer gleich ist, weil den Kontakt zwischen der Haut und der Oberfläche immer anders ist. Zum Beispiel auf dem Bett, bzw. unter ihrem Finger ist 1g Isotop, doch nur 0.1g geht auf die haut... ergo pt=0.1g/1g. Jedes mal wenn Sie was anders anfassen ist pt anders. So for each surface contact we have a different $pt$.2012-08-12