Location shift of two distributions

Question

Consider two independent random variables $X$ and $Y$. Let $\mathbb{Y}$ be the support of $Y$ and take a function $f:\mathbb{Y}\rightarrow \mathbb{R}$. Consider the distribution of $$ X+f(Y) \text{ conditional on $Y=y$} $$ and the distribution of $$ X+f(Y) \text{ conditional on $Y=y'$} $$ Is it correct to say that these two distributions will have all equal central moments except the mean?

This is correct when $X$ has, for instance, a Normal distribution. However, I am not sure if it holds generically.

Answer 1

Yes, the conditional higher-order central moments are all the same

Since $X$ and $Y$ are independent, the conditional expectation has the following form:

$$\begin{align} \mu(y) &\equiv \mathbb{E}[X+f(Y) | Y=y] \\[6pt] &= \mathbb{E}[X | Y=y] + \mathbb{E}[f(Y) | Y=y] \\[6pt] &= \mathbb{E}[X] + f(y), \\[6pt] \end{align}$$

Using the independence condition again, you then have conditional higher-order central moments given by:

$$\begin{align} \mathbb{E}[(X+f(Y) - \mu(y))^k | Y=y] &= \mathbb{E}[(X+f(Y) - \mathbb{E}[X] - f(y))^k | Y=y] \\[6pt] &= \mathbb{E}[(X+f(y) - \mathbb{E}[X] - f(y))^k | Y=y] \\[6pt] &= \mathbb{E}[(X - \mathbb{E}[X])^k | Y=y] \\[6pt] &= \mathbb{E}[(X - \mathbb{E}[X])^k], \\[6pt] \end{align}$$

which is just the $k$th higher-order central moment of $X$. As you can see, this does not depend on the conditioning value $y$, so the conditional higher-order central moments are all the same regardless of the conditioning value.

Answer 2

If $f$ is identical function:

We want to find: $$\mathbb{E}[(X + Y - \mathbb{E}[X + Y])^n \mid Y = y]$$

Since $X,Y$ are independent, $$\mathbb{E}[(X + Y - \mathbb{E}[X + Y])^n \mid Y = y]=\mathbb{E}[(X + Y - (\mathbb{E}[X] + \mathbb{E}[Y]))^n \mid Y = y]$$

rewrite it as: $$\mathbb{E}[(X - \mathbb{E}[X]) + (Y - \mathbb{E}[Y]))^n \mid Y = y]$$

Using binomial theorem: $$ (x+y)^n=\binom{n}{0} x^n y^0+\binom{n}{1} x^{n-1} y^1+\binom{n}{2} x^{n-2} y^2+\cdots+\binom{n}{n-1} x^1 y^{n-1}+\binom{n}{n} x^0 y^n $$

when $k \neq 0$ $$\mathbb{E}[\binom{n}{k} (X - \mathbb{E}[X])^{n-k} (Y - \mathbb{E}[Y])^k|Y=y]=\binom{n}{k} (X - \mathbb{E}[X])^{n-k} \mathbb{E}[(Y - \mathbb{E}[Y])^k|Y=y]=\binom{n}{k} (X - \mathbb{E}[X])^{n-k}\times 0=0$$

Thus $$\mathbb{E}[(X - \mathbb{E}[X]) + (Y - \mathbb{E}[Y]))^n \mid Y = y]=\mathbb{E}[\binom{n}{n} (X - \mathbb{E}[X])^n(Y - \mathbb{E}[Y])^0|Y=y]=\mathbb{E}[(X - \mathbb{E}[X])^n|Y=y]=(X - \mathbb{E}[X])^n$$

If $f$ is any other function, the results are the same.