Mean of the Exponential Martingale

Since $\theta$ is a constant, $\mathbb{E}[Z] = e^{-\theta^2/2}\,\mathbb{E}[e^{\theta X}]$ , so it is enough to compute the moment generating function of $X$ . Complete the square in the exponent:

\theta x - \tfrac{x^2}{2} = -\tfrac12 (x-\theta)^2 + \tfrac{\theta^2}{2}.

Then

\mathbb{E}[e^{\theta X}] = \int_{-\infty}^{\infty} e^{\theta x}\,\frac{1}{\sqrt{2\pi}}e^{-x^2/2}\,dx = e^{\theta^2/2}\int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}}e^{-(x-\theta)^2/2}\,dx = e^{\theta^2/2}.

The last integral equals $1$ because its integrand is the $\mathcal{N}(\theta,1)$ density. Therefore

\mathbb{E}[Z] = e^{-\theta^2/2}\,e^{\theta^2/2} = 1.

\boxed{\mathbb{E}[Z] = 1 \text{ for every } \theta\in\mathbb{R}}

Key idea. Multiplying the $\mathcal{N}(0,1)$ density by $e^{\theta x-\theta^2/2}$ gives exactly the $\mathcal{N}(\theta,1)$ density. So $Z$ is a likelihood ratio, and its mean is $1$ . This is the one-step version of the exponential martingale $e^{\theta W_t-\theta^2 t/2}$ and of the Girsanov change of measure.