# I have distribution that range from 0-1.8 and mean E[X] I want to move the mean by y How I modify original distribution so that it has mean E[X] -y?

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?

• Do you mean that the transformed distribution must still have support $(0,1.8)?$ – BruceET Nov 4 '20 at 8:41
• yes, That is what I want. After transformation, Distribution range still should be between o to 1.8 – ASD Nov 4 '20 at 9:04
• Replacing it by any distribution with mean $E[X]-y$ is the most general solution. – whuber Nov 4 '20 at 15:01

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 $$\mathbb P(0 \le Z \le 1.8)=1$$
• @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 $X$. – Henry Nov 4 '20 at 20:44