I have a trouble with this problem involving Hermite polynomial(probability version!). The problem is
$$ \frac {(-1)^{r-1}H_{2r-1}(x)}{2^{r-1}(r-1)!x}=\sum_{s=0}^{r-1}\frac{(-1)^s}{2^ss!}H_{2s}(x) $$
Some paper (Wand, Schucany 1990) said the above equality is satisfied if using recurrence relation,
$$ H_j(x)-xH_{j-1}(x)+(j-1)H_{j-2}(x)=0 $$ I tried to solve it as best could, but I don't make it. Please, help!
Through the recurrence relation we have: $$ \frac{H_{2r-1}}{x} = H_{2r-2}-(2r-2)\frac{H_{2r-3}}{x}\tag{1} $$ so: $$ \frac{H_{2r-1}}{x} = H_{2r-2}-(2r-2)H_{2r-4}+(2r-2)(2r-4)\frac{H_{2r-5}}{x}\tag{2} $$ and by induction (since $\frac{H_1}{x}=H_0$): $$ \frac{H_{2r-2}}{x}=\sum_{k=0}^{r}(-1)^k H_{2r-2k}\prod_{j=1}^{k}(2r-2j) \tag{3}$$ that is easy to rearrange in the wanted form.