$\int_{0}^1W_x(t)dW_y(t)/(\int_{0}^1W_x^2(t)dt)^{1/2}$ normally-distributed?

I have came across the following stochastic integrals: $$\frac{\int_{0}^1W_x(t)dW_y(t)}{(\int_{0}^1W_x^2(t)dt)^{1/2}}$$ which was claimed to be standard normally distributed ($W_x$ and $W_y$ are independent standard Wiener processes). I simulated it and it is clearly the case, but I cannot show it formally. Any suggestions?

Thanks, Tamas

• are you sure you don't miss an expectation in the denominator? It should help to use the properties of the Itô integral $I_t = \int_0^t W^x_u dW^y_u$ (mean is zero and variance given by Itô isometry) and then specify to the case $t=1$. – Quantuple Apr 11 '18 at 15:27
• Any background information for this question? – Gordon Apr 12 '18 at 19:58