0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Do you mean that the transformed distribution must still have support $(0,1.8)?$ $\endgroup$ – BruceET Nov 4 '20 at 8:41
  • $\begingroup$ 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
  • $\begingroup$ Replacing it by any distribution with mean $E[X]-y$ is the most general solution. $\endgroup$ – whuber Nov 4 '20 at 15:01
1
$\begingroup$

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 $\mathbb P(0 \le Z \le 1.8)=1$

$\endgroup$
  • $\begingroup$ 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
  • $\begingroup$ @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
  • $\begingroup$ Yes, It is continuous distribution between 0 and 1.8 and had point mass at 0. $\endgroup$ – ASD Nov 5 '20 at 7:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.