Anybody see this kind of DC restorer or clamper, would you like to give out the detailed solution for the output waveform
You can solve this pretty easily. Once steady-state arrives, you know that the currents flowing in and out must be the same. Early on, the voltage across the capacitor is zero. But the capacitor must acquire a mean voltage across it that results in equal currents flowing in one direction matching the currents flowing in the opposite direction when it is in steady-state.

So, it follows that:

$$\frac1{\pi}\int_0^\pi \frac{V_0\operatorname{sin}\theta-V_\text{D}+V_\text{C}}{R_2}\:\text{d}\theta+\frac1{\pi}\int_0^\pi \frac{V_0\operatorname{sin}\theta+V_\text{C}}{R_1}\:\text{d}\theta+\frac1{\pi}\int_0^\pi \frac{-V_0\operatorname{sin}\theta+V_\text{C}}{R_1}\:\text{d}\theta=0\:\text{A}$$

This solves out as:

$$V_\text{C}=\frac{V_\text{D}-\frac{2\cdot V_0}{\pi}}{1+2\frac{R_2}{R_1}}$$

Given the low currents in the diode, indicated by \$R_1=R_2=100\:\text{k}\Omega\$, I might estimate \$V_\text{D}\approx 400\:\text{mV}\$. From this, I find that \$V_\text{C}\approx -1.989\:\text{V}\$. Which explains the approximate \$2\:\text{V}\$ shift in your output.

Please note that the above assumes that the diode's conduction angle is equal to the angle when it doesn't conduct. This is "mostly true" when \$R_1\approx R_2\$. But if you change the value of \$R_2\$ so that the conduction angle changes significantly, the above formula isn't correct. (I assumed a simplification to get an easy result.)

But since you only asked to explain the result you have, the above is enough.