Cfrgtkky

I was given that that $f_{X|Y}(x|y=1)=lambda_1e^{-lambda_1 x}$ for $xgeq 0$ and $f_{X|Y}(x|y=-1)=lambda_2e^{lambda_2 x}$ for $xleq 0$. $X$ is continuous. $Y$ is discrete and can only take on those two values, $1$ and $-1$, with probability $p$ and $1-p$ respectively. $lambda_1$ and $lambda_2$ are positive values. I am trying to get the marginal PDF, $f_x(x)$ for when $xgt 0$ and for when $xlt 0 $.

I got $f_x(x)$ = $sum_{y}P(Y)f_{X|Y}(x|y) = plambda_1e^{-lambda_1 x} +(1-p)lambda_2e^{lambda_2 x}$, but after this , I am not sure how to interpret it for $xgt 0$ and $xlt 0 $. If I plot $f_{X|Y}(x|1)=lambda_1e^{-lambda_1 x}$ for $xgeq 0$, $f_{X|Y}(x|-1)=lambda_2e^{lambda_2 x}$ for $xleq 0$, and $lambda_1e^{-lambda_1 x} +lambda_2e^{lambda_2 x}$ on the same graph, I obtain the following:
plot for a fixed $lambda$, but even visualizing it doesn't help me understand it any better.

To better keep track of the conditional statements you can rewrite the conditional densities as functions over $mathbb{R}$ using indicator functions, so
$$
f_{X|Y}(x|y=1) = lambda_1e^{-lambda_1 x}mathbb{1}_{x geq 0}, qquad f_{X|Y}(x|y=-1)=lambda_2e^{lambda_2 x}mathbb{1}_{x < 0},
$$
we haven't added any new information, but it might help to keep track of everything. So now when you carry out your marginalisation you have
$$
begin{align}
f_X(x) &= sum_{y in {-1, 1}} f_Y(y)f_{X|Y}(x|y) \
&= plambda_1e^{-lambda_1 x}mathbb{1}_{x geq 0} + (1-p)lambda_2 e^{lambda_2 x}mathbb{1}_{x < 0},
end{align}
$$
again nothing new, but hopefully now it should be just a little clearer how to interpret it, so you could also rewrite the density as
$$
f_X(x) =
begin{cases}
plambda_1 e^{-lambda_1 x} & x geq 0 \
(1-p)lambda_2 e^{lambda_2 x} & mbox{otherwise}.
end{cases}
$$

So on the sets ${ x < 0 }$ and ${ x > 0 }$ you are going to have a graph that looks like the density function of an exponential random variable weighted by the probabilities $p$ or $(1-p)$ and you will have a discontinuity at $0$.

A good thing to do next for your own understanding is to probably rewrite the density function in terms of the absolute value of $x$, $| x |$, and see if you can realise the Laplace distribution as a special case of this setup.