# Find the MLE of $\mathbb{P}(X<Y)$ for $X\sim N(\mu_X,\sigma^2)$, $Y\sim N(\mu_Y,\sigma^2)$

I wish to find the MLE of $$\mathbb{P}(X after having calculated the MLE's of $$X\sim N(\mu_X,\sigma^2)$$, $$Y\sim N(\mu_Y,\sigma^2)$$ where $$\sigma^2$$ is known. We have i.i.d $$x_1,...x_n$$ and $$y_1,...,y_n$$ data from the respective distrabutions above.

I have calculated that $$\hat{\mu}_X = \overline{x}$$ and $$\hat{\mu}_Y = \overline{y}$$.

I also know about the invariance properties of maximum likelihood but I'm unsure how to encorporate it above. I know by symmetry that $$\mathbb{P}(X but from there?...

Thanks

• X, Y are independent, we assume? A neat trick is to write $V=X<Y$ as a Bernoulli event. Then $P(X<Y) = E(V)$. Then you can say $E(V) = E(E(V|Y=y))$ by the law of total expectation. May 17 at 21:03
• Use the fact that $X-Y$ is normal (assuming $X,Y$ are independent) to find $P(X-Y<0)$ in terms of the population parameters. Then use invariance of MLE. This probability would have been $1/2$ if $\mu_X=\mu_Y$ but then there would be no point estimating this quantity. May 17 at 21:59
• @StubbornAtom Okay, $Z \sim N(\mu_X - \mu_Y , 2\sigma^2)$, so if $Z$ is normally distrabuted like you mentioned, then the MLE of $P(Z<0)$ is $\Phi(\frac{-\mu_Z}{\sqrt{2\hat{\sigma}^2}}) = 1 - \Phi(\frac{\mu_Z}{\sqrt{2\hat{\sigma}^2}})$ where $\Phi$ is the standard Normal CDF, which can be done due to the invariance property? May 18 at 11:22
• Almost. $\mu_Z$ has to be replaced by its MLE; $\sigma^2$ is known, so doesn't require estimation. As you have solved this, you could post a detailed answer below. May 18 at 11:45
• If these are paired data, apply stats.stackexchange.com/questions/511265 to the variable $X-Y.$
– whuber
May 18 at 12:21

We know $$X-Y$$ is normally distributed so let $$Z=X-Y$$ then, $$$$Z\sim N(\mu_X - \mu_Y,2\sigma^2)$$$$ MLE of $$\mu_X$$ and $$\mu_Y$$ are $$\hat{\mu}_X = \overline{x}$$ and $$\hat{\mu}_Y = \overline{y}$$, which are the corresponding sample means. Then using above, the MLE of $$\mu_Z=\mu_X-\mu_Y$$ is clearly $$\overline{x}-\overline{y}$$ and we'll call it $$\hat{\mu}_Z$$.

Using the substitution above,
$$$$P(X

Now let $$\Phi(z)$$ represent the standard normal CDF, then the MLE of $$P(Z<0)$$ is,

$$$$\Phi\bigg(\frac{-\hat{\mu}_Z}{\sqrt{2\sigma^2}}\bigg) = 1 - \Phi\bigg(\frac{\hat{\mu}_Z}{\sqrt{2\sigma^2}}\bigg) = 1 - \Phi \bigg(\frac{\overline{x}-\overline{y}}{\sqrt{2\sigma^2}}\bigg)$$$$

which we can do since the MLE is invariant against coordinate transformations.

• $\hat \mu_X$ is $\overline x$ and likewise $\hat \mu_Y$ is $\overline y$. So if you let $\mu_Z=\mu_X-\mu_Y$, then $\hat\mu_Z=\overline x-\overline y$. May 18 at 19:01

It seems you need a simple z-test of $$H_0: \mu_x=\mu_y$$ against $$H_a: \mu_x<\mu_y,$$ where the z-statistic is $$Z = \frac{\bar X-\bar Y}{\sigma\sqrt{2/n}},$$ rejecting at level 5% if $$Z < -1.645.$$

Let $$n = 20, \sigma = 5, \mu_x = 45, \mu_y = 50.$$ Then data might be similar to the fictitious data sampled and summarized below in R.

set.seed(517)
n = 20
x = rnorm(n, 45, 5)
y = rnorm(n, 50, 5)

summary(x); length(x); sd(x)
Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
29.43   40.92   43.85   43.79   46.75   56.61
[1] 20          # sample size
[1] 5.943466    # sample SD
summary(y); length(y); sd(y)
Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
40.21   47.51   51.91   51.59   56.63   58.59
[1] 20
[1] 5.345866
boxplot(x,y, col="skyblue2")


Then the z-statistic $$Z < -1.645$$ is computed as shown below. The null hypothesis is rejected according to the criterion mentioned above. Also, he P-value of the test is near $$0.$$

se = 5*sqrt(2/n)
z = (mean(x)-mean(y))/se
z
[1] -4.933073       # z-statistic
pnorm(z)
[1] 4.047292e-07    # P-value


If there is any doubt that $$P(X < Y) > 1/2,$$ that is, $$Y$$ stochastically dominates $$X,$$ then consider the plots of the empirical CDFs (ECDFs) of the samples x and y. Values sampled from $$Y$$ (blue) tend to be larger: They plot to the right of (thus below) values sampled from $$X.$$

hdr="ECDFs of Samples from Y (blue) and X"
plot(ecdf(y), col="blue", main=hdr)
lines(ecdf(x), col="brown")


Addendum: Explicit MLE of $$P(X < Y)$$ based on my fictitious data:

The MLEs of $$\mu_x, \mu_y$$ are $$\bar X, \bar Y,$$ respectively. By invariance, the MLE of $$\mu_x-\mu_y$$ is $$\bar X - \bar Y = -7.8.$$ The MLE of $$P(X < Y)$$ can be found by standardizing $$P(X < Y) = P(X -Y = 0),$$ using MLEs for parameters.

$$P(X - Y < 0) = P\left(\frac{(X-Y)-(\hat \mu_x-\hat \mu_y)}{\sigma\sqrt{2}} < \frac{\hat \mu_y-\hat\mu_x}{\sigma\sqrt{2}}\right)\\ =P\left(Z < \frac{7.8}{7.071} = 1.103\right) = 0.8650,$$ where $$Z$$ is standard normal.

The exact value, based on parameters (rather than their MLEs from samples of size $$20)$$ is $$P(X < Y) = 0.7602.$$

pnorm(5/(5*sqrt(2)))
[1] 0.7602499


Simulation (3 place accuracy):

set.seed(519)
X = rnorm(10^7, 45,5)
Y = rnorm(10^7, 50,5)
mean(X < Y)
[1] 0.7602937


The value from MLEs above can also be approximated by simulation.

set.seed(520)
X = rnorm(10^7, 43.79, 5)
Y = rnorm(10^7, 51.59, 5)
mean(X < Y)
[1] 0.8648917

• Appreciate this however was looking for a more theoretical approach, maybe should have made it clearer in the question. May 18 at 9:38
• Sorry but how does this answer the question? May 18 at 10:22
• Obvious comments to you sir, not to OP. These several obvious and cryptic comments have answered the question, thankfully. May 19 at 6:21
• Addendum with explicit MLE for $P(X < Y).$ May 19 at 17:22