# If $T(\bf{X})$ is a sufficient statistic for $\theta$, why does the conditional distribution of $\bf{X}$ given $T(\bf{X})$ doesn't depend on $\theta$? [duplicate]

I know that if $T(\bf{X})$ is a sufficient statistic for $\theta$, then the conditional distribution of $\bf{X}$ given $T(\bf{X})$ doesn't depend on $\theta$. However, I am not sure why this makes sense.

It seems that we will never know $\theta$, and that $\bf{X}$ to begin with already depends on $\theta$, and that $T(\bf{X})$ is a "coarser" summary than $\bf{X}$, so how can it be that given $T(\bf{X})$, the distribution of $\bf{X}$ no longer depends on $\theta$? Is there an intuition here?

• When I know $T(x)$, I no longer need to know the value of $\theta$ to compute the probability of $X$. It's just sufficient to know the value of $T(x)$ Dec 3 '16 at 21:35
• What is your definition of "sufficient statistic"? One standard definition of sufficient statistic is "A statistic $t=T(X)$ is sufficient for underlying parameter $\theta$ precisely if the conditional probability distribution of the data $X$, given the statistic $t=T(X)$, does not depend on the parameter $\theta$." If this is also your definition, then what you are asking is why this definition makes sense. Dec 4 '16 at 20:28