[Math] Probability of a point from one normal distribution being higher than a point from another independent normal distribution

normal distributionprobability distributionsstatistics

Given two independent normal distributions:

Distribution 1: Mean $= 23.95$, SD $= 7.44$

Distribution 2: Mean $= 16.29$, SD $= 7.79$

How often on average will a point from Distribution 2 be greater than a point from Distribution 1?

I apologize for any nomenclature that is incorrect.

Progress

I know that the distribution of $Z=Y-X$ is normal, as well as its mean and variance.

Best Answer

Let $ X_1 \sim N(23.95,7.44), X_2 \sim N(16.29,7.79) $,

then, as mentioned by Dilip Sarwate in the comments, $ X_1 - X_2 \sim N(7.66,{\sqrt{7.44^2 + 7.79^2}}) $

Thus the problem becomes $$\mathbb{P}(X_1<X_2) = \mathbb{P}(X<0),\ \ where \ \ X = X_1 - X_2 $$

This can be calculated directly or reverting to the standard $ N(0,1) $ distribution as follows:

$$ \begin{align} \mathbb{P}( X < 0 ) = \mathbb{P}(\frac{X-7.66}{{\sqrt{7.44^2 + 7.79^2}}}<\frac{-7.66}{\sqrt{7.44^2 + 7.79^2}}) \end{align} $$

Noting that $$ \frac{X-7.66}{{\sqrt{7.44^2 + 7.79^2}}} = Z \sim N(0,1) $$

So the required probability is, approximately,

$$ \begin{align} \mathbb{P}(Z<-0.7110976012) &= \mathbb{P}( Z > 0.7110976012) \\ &= 1 -\mathbb{P}(Z<0.7110976012) \\ &\approx 1 - 0.7611\\ &= 0.2389 \end{align} $$

The numerical value found by looking up 0.71 in a standard normal table.

Related Solutions

Probability – Point from Normal Distribution Greater Than Another Point

Suppose that $X_1 \sim {\rm N}(\mu_1,\sigma_1^2)$ and $X_2 \sim {\rm N}(\mu_2,\sigma_2^2)$ are independent. Then, $$ {\rm P}(X_1 > X_2 ) = {\rm P}(X_1 - X_2 > 0) = 1 - {\rm P}(X_1 - X_2 \le 0). $$ Now, by independence, $X_1 - X_2$ is normally distributed with mean $$ \mu := {\rm E}(X_1 - X_2) = \mu_1 - \mu_2 $$ and variance $$ \sigma^2 := {\rm Var}(X_1 - X_2) = \sigma_1^2 + \sigma_2^2. $$ Hence, $$ \frac{{X_1 - X_2 - \mu}}{{\sigma}} \sim {\rm N}(0,1), $$ and so $$ {\rm P}(X_1 - X_2 \le 0) = {\rm P}\bigg(\frac{{X_1 - X_2 - \mu }}{\sigma } \le \frac{{0 - \mu }}{\sigma }\bigg) = \Phi \Big( \frac{-\mu }{\sigma }\Big), $$ where $\Phi$ is the distribution function of the ${\rm N}(0,1)$ distribution. Thus, $$ {\rm P}(X_1 > X_2 ) = 1 - {\rm P}(X_1 - X_2 \le 0) = 1 - \Phi \Big( \frac{-\mu }{\sigma }\Big). $$

[Math] Distribution of largest sample from normal distribution.

The random variable $Z = \max_{i=1}^n(X_i)$ is known as order statistics, and is sometimes denoted as $X_{n:n}$.

The cumulative density function of $Z$ is easy to find: $$ F_Z(z) = \mathbb{P}\left(Z \leqslant z\right) = \mathbb{P}\left(\max_{i=1}^n(X_i) \leqslant z\right) = \mathbb{P}\left( X_1 \leqslant z, X_2 \leqslant z, \ldots, X_n \leqslant z\right) $$ using independence: $$ F_Z(z) = \left(F_X(z)\right)^n $$ Thus the density function is $$ f_Z(z) = n f_X(z) F_X^{n-1}(z) $$ In particular, it follows that $Z$ is not normal.

Expected values of $Z$ are known in closed form for $n=1,2,3,4,5$ (asking Mathematica):

In[31]:= Table[
 Mean[OrderDistribution[{NormalDistribution[m, s], n}, n]], {n, 1, 5}]

Out[31]= {m, m + s/Sqrt[Pi], m + (3 s)/(2 Sqrt[Pi]), 
 m + (6 s ArcTan[Sqrt[2]])/Pi^(3/2), 
 m - (5 s)/(2 Sqrt[Pi]) + (15 s ArcTan[Sqrt[2]])/Pi^(3/2)}

Large $n$ asymptotics is discussed here.

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Best Answer

Related Solutions

Probability – Point from Normal Distribution Greater Than Another Point

[Math] Distribution of largest sample from normal distribution.

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