$Y$ is normally distributed or constant if $X$ and $X+Y$ are normally distributed and $X$, $Y$ are independent

Question

Let $X$, $Y$ be independent random variables with $X$ and $X+Y$ normally distributed. Prove that $Y$ is also normally distributed or is constant a.s.

Answer 1

Hint: consider the characteristic function $\phi_Y(t) = \mathbb E[\exp(itY)]$.

Answer 2

By independence $$φ_{X+Y}(t)=φ_X(t)φ_Y(t)$$ where $φ$ denotes the characteristic function. Hence, since $X,X+Y$ are normally distributed, with parameters (say) $X\sim N(μ_X,σ^2_X)$ and $X+Y\sim N(μ_1,σ^2_1)$ you have that $$e^{itμ_1-\frac12σ_1^2t^2}=e^{itμ_X-\frac12σ_X^2t^2}φ_Y(t)\implies φ_Y(t)=e^{it(μ_1-μ_x)-\frac12t^2(σ^2_1-σ^2_X)}$$ hence $Y$ is also normally distributed with parameters $μ_Y=μ_1-μ_Χ$ and $σ^2_Y=σ^2_1-σ^2_X$. Note that a constant random variable $Y\equiv c$ is also normally distributed with mean $c$ and variance $0$ (so, the two cases in the exercise are not really necessary imho).

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As pointed out in the comments: $σ_1$ must be $\ge σ_Χ$. By independence this is indeed the case here, since $$σ_1=Var(X+Y)=Var(X)+Var(Y)\ge Var(X)=σ_X$$