Rao-Blackwellisation using non-sufficient statistics The following is given as a remark in chapter 7 of Introduction to mathematical statistics Hogg and Craig, 8th edition. (It is mentioned as "Remark 7.3.1")

Now, I do understand that the distribution of $Y_1|Y_3$ would contain $\theta$ compulsorily (because $Y_3$ is not a sufficient statistic).
But I do not understand why $\theta $ has to be present compulsorily in the conditional mean $X=\mathbb E[\varphi(Y_1)|Y_3]$.
I can agree to the statement "$\theta$ can be present in $X$". But I do not understand why $\theta$ has to be present in $X$.
Because the distribution of $Y_1|Y_3$ could contain parameters other than $\theta$ also. So, it is possible for $\mathbb E[\varphi(Y_1)|Y_3]=X$ to contain those other parameters(and not contain $\theta$).
So why is it always true that $\theta $ will be present in $X$?
EDIT:-
I'm not sure, but I think that (going by the definition of sufficiency given in Hogg and Craig) the concept of sufficiency is defined only when the pdf of the random sample ( i.e. $f_X(x;\theta)$ ) has exactly one unknown parameter (namely $\theta$).
If that's the case, then the only unknown parameter that the distribution of $Y_1|Y_3$ contains is $\theta$. 
Also note that $X=\mathbb E[\varphi(Y_1)|Y_3]$, cannot be a constant that is not equal to $\theta$( because then $\mathbb{E}[X]\neq \theta$ )
So, If $X$ is a constant then $X=\theta$. But what if $X$ is not a constant?
So, in the case when $X$ is not a constant, why should $\theta$ be present in $X$, compulsorily?

Counter-example: If one considers the case of a sample $(X_1,\ldots,X_n)$ from a $N(\theta,\theta^2)$ distribution, a minimal sufficient statistic is
$$T=\left(\sum_i X_i,\sum_i(X_i-\bar X_n)^2\right)$$
Now $\varphi(T)=\bar X_n$ is an unbiased estimator of $\theta$ that is not sufficient. But the conditional expectation
$$\mathbb E_\theta[\varphi(T)|\varphi(T)]=\varphi(T)$$
does not involve $\theta$.
