An algebraic manipulation I can't get

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I'm working with present value models. I don't get this passage: $$ R_t = \theta_t + \frac{1}{h}\sum_{j=0}^{h-1}\mathbb{E_t}r_{t+j} $$ Where $\mathbb{E_t}$ denotes the expectation conditioned on $t$, and $$ \theta_t := \frac{1}{h}\log{\Theta_t} $$ Where $\Theta_t$ is a costant. From where I'm studying, it says that subtracting $r_t$ from both sides yields:

$$ R_t - r_t = \sum_{j=1}^{h-1}\left(1-\frac{j}{h}\right)\mathbb{E_t}\Delta r_{t+j} + \theta_t $$ Where $\Delta$ is the difference operator $\Delta r_t := r_t - r_{t-1}$. I don't know which manipulation I should do after I subtract the quantity from both sides in order to get the above result. Can anybody please help me?

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It seems to me that we must show that

$$\frac 1 h [\sum_{j=0}^{h-1} E_t r_{t+j} - hr_t] = \sum_{j=0}^{h-1} (1-\frac{j}{h})E_t\Delta r_{t+j}$$

What I tried:

$$RHS = \sum_{j=0}^{h-1} (\frac{h-j}{h})E_t\Delta r_{t+j}$$

$$ = \frac{1}{h}\sum_{j=0}^{h-1} (h-j)E_t\Delta r_{t+j}$$

$$ \stackrel{(*)}{=} \frac{1}{h}[\sum_{j=1}^{h-1} E_t r_{t+j} - (h-1)E_tr_t] \stackrel{(**)}{=} LHS$$

$(*)$ Convince yourself.

$(**)$ $E_t r_t = r_t$ because $E_t r_{t+j} = E[r_{t+j} | r_0, r_1, ..., r_t]$ ?