The joint density of $X$ and $Y$ is given by : $f(x,y)=\frac{1}{\sqrt{2 \pi}}e^{-y}e^{-\frac{(x-y)^2}{2}}$. The joint density of $X$ and $Y$ is given by
  $$f\left(x,y\right)=\frac{1}{\sqrt{2\pi}}e^{-y}e^{-\frac{\left(x-y
\right)^{2}}{2}}$$
  Here $0&lt;y&lt;\infty$ and $-\infty&lt;x&lt;\infty$. Compute the individual MGFs of $X$ and $Y$.


I could find the marginal distribution of $Y$ which is ${\rm Exp}\left(1\right)$, so I can find its MGF. But I cannot find the marginal distribution of $X$. Please help!
Assuming "MGF" means "moment generating function":

In this set-up, $Y$ is exponential and conditional on $Y$, the rv. $X$ is $N(Y,1)$.  That is, $X$ can be represented as $Y+Z$, where $Z\sim N(0,1)$ is independent of $Y$.  So the MGF of $X$ is $E\exp (uX) = E\exp(uY+uZ) = E\exp(uY)\exp(uZ)=\frac 1{1-u}\exp(u^2/2)$.  The MGF of $Y$ is $E\exp(vY)=1/(1-v)$.
You don't need to find the distribution of $X$ to derive its MGF.

As @kimchilover has already mentioned, from the joint pdf of $(X,Y)$ it is clear that $X$ conditioned on $Y=y$ has a $\mathcal{N}(y,1)$ distribution, where $Y$ itself is distributed as $\text{Exp}(1)$. 

MGF of $X$ is

$$M_X(t)=E\,[e^{tX}]=E\,E\,[e^{tX}\mid Y]=E\,[e^{Yt+t^2/2}]=e^{t^2/2}E\,[e^{tY}]$$

The only fact used above is the MGF of a univariate Normal distribution.

Now MGF of $Y$ is $E\,[e^{tY}]=\frac{1}{1-t}$ for $t&lt;1$.

Therefore the required MGF is $$M_X(t)=\frac{e^{t^2/2}}{1-t}\quad,\,t&lt;1$$