Probability normal distribution P(X>Y)?

Question

Angelo earns every month as a variable normal X N(1000;400^2), Bruno N(1400;300^2). Calculate the probability of Angelo earns more then Bruno p(X>y)?

Answer 1

Hint: Compute this conditionally to the value of $X$. Then take the expectation.

Answer 2

Hint:

Prescribe $f$ by $(x,y)\mapsto 1$ if $x>y$ and $(x,y)\mapsto 0$ otherwise. Then calculate $\mathbb Ef(X,Y)$.

Answer 3

This problem can be decomposed into

$$ \begin{align} Pr(X = x \;\&\; Y > x) =& Pr(X = x) \times Pr(Y > x) \;\; \forall x \in \mathbb{R}\\ =& f_{\mathbf{X}}(x) \times (1-F_{\mathbf{Y}}(x)) \;\; \forall x \in \mathbb{R} \end{align} $$

where $\mathbf{X}$ is the income of Angelo and $\mathbf{Y}$, the one of Bruno, $f_{\mathbf{X}}(x)$ is the density of $\mathbf{X}$, $F_{\mathbf{Y}}(x)$ is the cumulatve distribution function, and assuming independence of $\mathbf{X}$ and $\mathbf{Y}$.

You get rid of the $\forall x$ by integrating out $x$, with

$$ \int_{-\infty} ^{\infty} f_{\mathbf{X}}(x) (1-F_{\mathbf{Y}}(x)) \, \mathrm{d} x$$

yet, this formula has no closed-form solution. Using numerical integration, you should find a numerical approximation, something around 78.8%.

An alternative approach is to compute the random variable $\mathbf W= \mathbf X-\mathbf Y$. As it is well known, the sum of two normally distributed random variable is also normally distributed with mean the sum of the mean (but one is negative) and the variance the sum of the individual variances.

Then, you need $1-F_{\mathbf W}(0)$ which is again $\approx $78.8%.