In a way, I want to change the original distribution so that it has desired mean of E[X]-y without shifting the distribution by y amount. Is there a way to do this?
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$\begingroup$ Do you mean that the transformed distribution must still have support $(0,1.8)?$ $\endgroup$ – BruceET Nov 4 '20 at 8:41
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$\begingroup$ yes, That is what I want. After transformation, Distribution range still should be between o to 1.8 $\endgroup$ – ASD Nov 4 '20 at 9:04
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$\begingroup$ Replacing it by any distribution with mean $E[X]-y$ is the most general solution. $\endgroup$ – whuber♦ Nov 4 '20 at 15:01
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One possibility is $$Z=X\left(1-\frac{y}{\mathbb E[X]}\right)$$
In that case you will have $$\mathbb E[Z] = \mathbb E[X]-y$$ and, if $\mathbb P(0 \le X \le 1.8)=1$ and $0 \lt y \le \mathbb E[X]$, then $\mathbb P(0 \le Z \le 1.8)=1$
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$\begingroup$ This is useful. but can it be possible to have the same range as 0≤X≤1.8. In this case, some point in right tail will have zero probability depending on y and E[X] Value. For example, if y is 0.1 and E[X] is 0.5 then 0.1/0.5=0.2 So after 1.8*(1-0.2)=1.44 all the x will have zero probability. will it be possible to have >0 probability for all the X between 0 to 1.8? Thanks. $\endgroup$ – ASD Nov 4 '20 at 16:52
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$\begingroup$ @AkshayChothani Yes and no. No for example if $X$ has probability $\frac12$ of being $1.8$ and probability $\frac12$ of being $0$; you cannot reduce the expectation without lowering the top end of the range. Yes if $X$ has a continuous distribution supported on $[0,1.8]$ (and some discrete distributions) as you might use a non-linear transformation specific to the distribution of $X$. $\endgroup$ – Henry Nov 4 '20 at 20:44
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$\begingroup$ Yes, It is continuous distribution between 0 and 1.8 and had point mass at 0. $\endgroup$ – ASD Nov 5 '20 at 7:24
