# If $X,Y$ are random variables and $Y\sim\mathcal N(x,\sigma^2)$ if $X=x$, are we able to conclude $Y-X\sim\mathcal N(0,\sigma^2)$?

Let $$\sigma>0$$ and $$Q(x,\;\cdot\;):=\mathcal N(x,\sigma^2)\;\;\;\text{for }x\in\mathbb R.$$ Note that $$Q$$ is a Markov kernel on $$(\mathbb R,\mathcal B(\mathbb R))$$.

Now, let $$X,Y$$ be real-valued random variables on a common probability space $$(\Omega,\mathcal A,\operatorname P)$$. Let $$X_\ast\operatorname P$$ and $$Y_\ast\operatorname P$$ denote the distribution of $$X$$ and $$Y$$, respectively, and asume that $$Y_\ast\operatorname P=(X_\ast\operatorname P)Q,$$ where the right-hand side denotes the composition of $$X_\ast\operatorname P$$ and $$Q$$.

Are we able to show that $$(Y-X)_\ast\operatorname P=\mathcal N(0,\sigma^2)$$?

Intuitively, if $$X=x\in\mathbb R$$, then $$Y\sim\mathcal N(x,\sigma^2)$$ and it's a well-known fact that $$Y-x\sim\mathcal N(0,\sigma^2)$$.

• Isn't the expected value of $Y-x$ equal to $-x$? – uniquesolution Mar 27 at 12:44
• @uniquesolution No, we should have $\operatorname E\left[Y\mid X=x\right]=x.$ – 0xbadf00d Mar 27 at 16:05
• Since the conditional distribution of $Y-X$ given $X=x$ is independent of $x$, then $Y-X$ is independent of $X$ and has this distribution (the particular distribution is not important). – zhoraster Mar 27 at 16:25
• @zhoraster Sorry, I think I don't understand what you mean. Clearly, a random variable is independent of a constant, but how do you conclude that $Y-X$ is independent of $X$? – 0xbadf00d Mar 27 at 16:33
• I mean this: two random variables $X$ and $Y$ are independent iff the conditional distribution $F_y$ of $X$ given $Y=y$ does not depend on $y$, i.e. $F_y\equiv F$. In this case also $F$ is the (unconditional) distribution of $X$. – zhoraster Mar 27 at 20:39

From the mentioned property of the normal distribution, we know that $$\int 1_B(y-x)\:\mathcal N(x,\sigma^2)({\rm d}y)=\mathcal N(0,\sigma^2)(B)\;\;\;\text{for all }(x,B)\in\mathbb R\times\mathcal B(\mathbb R)\tag1.$$ Now, it's easy to see that $$\operatorname P\left[(X,Y)\in\;\cdot\;\right]=X_\ast\operatorname P\otimes\:Q\tag2$$ and hence $$$$\begin{split}\operatorname P\left[Y-X\in B\right]&=\int\operatorname P\left[X\in{\rm d}x\right]\int Q(x,{\rm d}y)1_B(y-x)\\&=\int\operatorname P\left[X\in{\rm d}x\right]\int\mathcal N(0,\sigma^2)({\rm d}y)1_B(y)=\mathcal N(0,\sigma^2)(B)\end{split}\tag3$$$$ for all $$B\in\mathcal B(\mathbb R)$$.

Yes : We can prove that $$Y-X \sim \mathcal N( 0, \sigma^2)$$.

Proof : To see it, the key is to write law of $$Y-X$$ using the expectation of the conditional probability knowing $$X$$:

$$P(Y-X < t) = E( P( Y -X < t | X ) )$$

Then, notice that $$P( Y - X < t | X )$$ (which is a function of X) actually does not depend on $$X$$ :

Let $$N_a$$ denote a Gaussian variable $$N_a \sim \mathcal N( a, \sigma^2)$$ (not related to $$X$$ and $$Y$$). $$P( Y - X< t | X = x_0 ) = P( Y - x_0 < t | X = x_0 )$$ $$= P( N_{x_0} - x_0 < t )$$ $$= P( N_0 < t )$$ This last term is constant (i.e. does not depend on $$x_0$$), therefore its expectation is the same.

$$E( P( N_0 < t ) ) = P( N_0 < t )$$

We just have proven that $$P(Y-X < t) = P( N_0 < t )$$

Therefore, $$Y - X \sim N_0$$

• I didn't claim that $Y\sim\mathcal N(0,\sigma^2)$, but $Y-X\sim\mathcal N(0,\sigma^2)$. And that's obviously the case if $X$ is a constant. – 0xbadf00d Mar 27 at 16:06
• Indeed, my bad, I updated my answer. – Florian Mar 27 at 17:34