The sample median is an order statistic and has a non-normal distribution, so the joint finite-sample distribution of sample median and sample mean (which has a normal distribution) would not be bivariate normal. Resorting to approximations, asymptotically the following holds (see my answer here):
√n[(ˉXnYn)−(μv)]→LN[(00),Σ]
n−−√[(X¯nYn)−(μv)]→LN[(00),Σ]
with
Σ=(σ2E(|X−v|)[2f(v)]−1E(|X−v|)[2f(v)]−1[2f(v)]−2)
Σ=(σ2E(|X−v|)[2f(v)]−1E(|X−v|)[2f(v)]−1[2f(v)]−2)
where ˉXnX¯n is the sample mean and μμ the population mean, YnYn is the sample median and vv the population median, f()f() is the probability density of the random variables involved and σ2σ2 is the variance.
So approximately for large samples, their joint distribution is bivariate normal, so we have that
E(Yn∣ˉXn=ˉx)=v+ρσvσˉX(ˉx−μ)
E(Yn∣X¯n=x¯)=v+ρσvσX¯(x¯−μ)
where ρρ is the correlation coefficient.
Manipulating the asymptotic distribution to become the approximate large-sample joint distribution of sample mean and sample median (and not of the standardized quantities), we have
ρ=1nE(|X−v|)[2f(v)]−11nσ[2f(v)]−1=E(|X−v|)σ
ρ=1nE(|X−v|)[2f(v)]−11nσ[2f(v)]−1=E(|X−v|)σ
So
E(Yn∣ˉXn=ˉx)=v+E(|X−v|)σ[2f(v)]−1σ(ˉx−μ)
E(Yn∣X¯n=x¯)=v+E(|X−v|)σ[2f(v)]−1σ(x¯−μ)
We have that 2f(v)=2/σ√2π2f(v)=2/σ2π−−√ due to the symmetry of the normal density so we arrive at
E(Yn∣ˉXn=ˉx)=v+√π2E(|X−μσ|)(ˉx−μ)
E(Yn∣X¯n=x¯)=v+π2−−√E(∣∣∣X−μσ∣∣∣)(x¯−μ)
where we have used v=μv=μ. Now the standardized variable is a standard normal, so its absolute value is a half-normal distribution with expected value equal to √2/π2/π−−−√ (since the underlying variance is unity). So
E(Yn∣ˉXn=ˉx)=v+√π2√2π(ˉx−μ)=v+ˉx−μ=ˉx
E(Yn∣X¯n=x¯)=v+π2−−√2π−−√(x¯−μ)=v+x¯−μ=x¯