GATE 2018 EC Q.23: solution and answer

Official answer: 0.25Source: Official answer key, GATE 2018 EC (IIT Guwahati)

1. The four variables are independent and identically distributed, each $\mathcal{N}(0,1)$. Their joint density is $\varphi (x_1)\varphi (x_2)\varphi (x_3)\varphi (x_4)$, which stays the same if the variables are reordered. So every ordering of $X_1,\dots ,X_4$ is equally likely.

2. Each $X_i$ is continuous, so $P(X_i=X_j)=0$ for $i\ne j$. With probability 1 exactly one of the four is the smallest.

3. The events $A_i=\{X_i\text{ is the smallest}\}$, $i=1,\dots ,4$, are therefore mutually exclusive and exhaustive (up to a probability-zero set of ties): $$\sum _{i=1}^{4}P(A_i)=1$$

4. By symmetry (the variables are i.i.d.), each index is equally likely to be the minimum: $P(A_1)=P(A_2)=P(A_3)=P(A_4)$.

5. Hence $4P(A_4)=1\Rightarrow P(A_4)=\frac{1}{4}=0.25$.

6. Check by integration. Condition on $X_4=x$. By independence, $X_4$ is the smallest when each of $X_1,X_2,X_3$ exceeds $x$, which has probability $[1-\Phi (x){]}^3$: $$P(A_4)={\int }_{-\infty }^{\infty }\varphi (x)[1-\Phi (x){]}^{3}dx.$$ Put $u=1-\Phi (x)$, so $du=-\varphi (x)dx$. As $x$ runs from $-\infty$ to $\infty$, $u$ runs from $1$ to $0$: $$P(A_4)={\int }_1^0{u}^3(-du)={\int }_0^1{u}^{3}du=\frac{1}{4}.$$ No property specific to the normal distribution was used. The result holds for any four i.i.d. continuous random variables.