How to calculate the unconditional probability of a R.V. given the the probability of the R.V. conditioned on two events?

Question

Consider the following problem:

Let $X \sim \mathcal{N}(0,1)$. $Y$ is conditioned on a flip of a fair coin (independent from $X$). In the case of Heads, we let $Y=X$. In the case of Tails, we let $Y=−X$. Determine $f_Y$

Here's my attempt:

\begin{eqnarray} f_Y(y) &=& f_{Y|Heads}\mathbb{P}(Heads) + f_{Y|Tails}\mathbb{P}(Tails)\\ \end{eqnarray}

Considering the fact that $f_{Y|Tails} = - f_{Y|Heads}$

\begin{eqnarray} f_Y(y) &=& f_{Y|Heads} \left[ \mathbb{P}(Heads) - \mathbb{P}(Tails)\right]\\ f_Y(y) &=& f_X(x) \left[ \mathbb{P}(Heads) - \mathbb{P}(Tails)\right]\\ \end{eqnarray}

Since it is a fair coin, $\mathbb{P}(Heads) = \mathbb{P}(Tails)$, then

\begin{eqnarray} f_Y(y) &=& 0\\ \end{eqnarray}

However, I couldn't fully make sense of this solution. Have I made any mistakes throughout the resolution? If so, how could I have approached this problem differently?

Answer 1

$F_Y(t)=\Pr(Y\leq t)=\Pr(X\leq t)\Pr(H) + \Pr(-X\leq t)\Pr(T)=0.5[\Pr(X\leq t) + \Pr(X\geq -t)]$

From symmetry, $\Pr(X\geq -t)=\Pr(X\leq t)$ so $F_Y(t)=\Phi(t)$ which means that $Y\sim N(0,1)$.