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Find all random variables $X$ such that if $Y$ is $N(0,1)$ and independent from $X$, then $X+Y$ and $\frac{1}{3}X+2Y-1$ have the same distribution.

I use characteristic functions. $Y$ is $N(0,1)$ then $\phi_Y(t)=\exp(-\frac{t^2}{2})$. $X+Y$ and $\frac{1}{3}X+2Y-1$ need to have the same distribution so we have to have $\phi_{X+Y}(t)=\phi_{\frac{1}{3}X+2Y-1}(t)$. It's equivalent to:

$$\phi_X(t)\phi_Y(t)=\exp(-it)\phi_X(\frac{1}{3}t)\phi_Y(2t)$$ $$\phi_X(t)\exp(-\frac{t^2}{2})=\exp(-it)\phi_X(\frac{1}{3}t)\exp(-\frac{4t^2}{2})$$.

Can I get the general form for $X$ from this equation?

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Multiplying both sides with $e^{t^2/2}$ yields

$$\phi_X(t) = \phi_X \left( \frac{t}{3} \right) e^{-it} e^{-3t^2/2}. \tag{1}$$

By iterating $(1)$, we obtain

$$\phi_X(t) = \phi_X \left( \frac{t}{3^n} \right) \exp \left( -it \sum_{j=0}^{n-1} \frac{1}{3^j} - \frac{3 t^2}{2} \sum_{j=0}^{n-1} \frac{1}{3^{2j}} \right).$$

Letting $n \to \infty$ shows that $X$ is a Gaussian random variable and, moreover, we can also calculate its mean and variance explicitly.

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