Sample distribution of the minimum.
Consider a random sample \(X_1, \ldots,
X_n\) of continuous random variables, independent and identically
distributed with cumulative distribution function \(F\) and density \(f\). We define the sample minimum of these
\(n\) random variables as
\[
m_n := \min\{X_1, \ldots, X_n\}.
\]
Find a formula for \(P(m_n > t)\) in terms of \(F(t)\), with \(t > 0\).
Find the cumulative distribution function \(H_n\) of \(m_n\) and write its formula in terms of \(F\).
Construct the density function \(h_n\) of \(m_n\) and write its formula in terms of \(f\).
General Data
The probability of interest is \(P(m_n >
t)\), where \(m_n =
\min\{X_1,\ldots,X_n\}\).
The minimum is greater than \(t\) if
and only if every observation is greater than \(t\). In symbols,
\[ \{m_n > t\} = \{X_1 > t, X_2 > t, \ldots, X_n > t\}. \]
Since the random variables are independent, the probability factorizes as
\[ P(m_n > t) = P(X_1 > t, \ldots, X_n > t) = \prod_{i=1}^{n} P(X_i > t). \]
All factors are equal, because the \(X_i\) are identically distributed. For any \(i\),
\[ P(X_i > t) = 1 - F(t). \]
i.e.
\[ P(m_n > t) = [1 - F(t)]^n. \]
Consistency with the support of the distribution
The result above must agree with the natural behavior of the
distribution at the boundaries of its support.
If the support is \((a,b)\) (where
\(a\) and \(b\) can be finite or infinite), then:
Case 1: \(t \leq
a\).
In this region, \(F(t)=0\) because no
probability mass exists below the lower bound.
Substituting into the formula gives \[
P(m_n > t) = [1 - 0]^n = 1.
\]
This is reasonable: if \(t\) is smaller
than the minimum possible value of the distribution, then the sample
minimum is automatically greater than \(t\).
Case 2: \(t \geq
b\).
Here, \(F(t)=1\) since the distribution
places all its probability mass at or below the upper bound.
Substituting into the formula gives \[
P(m_n > t) = [1 - 1]^n = 0.
\]
This is also reasonable: if \(t\) is
larger than the maximum possible value of the distribution, it is
impossible for the sample minimum to exceed \(t\).
Thus, the expression \([1 - F(t)]^n\) not only provides the general formula but also behaves correctly at the endpoints of the support.
Final result:
\[ \boxed{P(m_n > t) = [1 - F(t)]^n} \]
The cumulative distribution function (CDF) of the sample minimum is defined as
\[ H_n(t) = P(m_n \leq t). \]
This can be expressed in terms of the complementary event:
\[ H_n(t) = 1 - P(m_n > t). \]
From part (a) it is known that
\[ P(m_n > t) = [1 - F(t)]^n. \]
Therefore,
\[ H_n(t) = 1 - [1 - F(t)]^n. \]
Consistency with the support of the distribution
As in part (a), the formula must agree with the natural behavior of the distribution at the boundaries:
Case 1: \(t \leq
a\).
For values below the lower bound, \(F(t)=0\). Substituting gives \[
H_n(t) = 1 - (1-0)^n = 0.
\]
This is expected: if \(t\) is smaller
than the minimum possible value of the distribution, the probability
that the sample minimum is less than or equal to \(t\) must be zero.
Case 2: \(t \geq
b\).
For values above the upper bound, \(F(t)=1\). Substituting gives \[
H_n(t) = 1 - (1-1)^n = 1.
\]
This is also consistent: if \(t\) is
larger than the maximum possible value of the distribution, the
probability that the sample minimum is less than or equal to \(t\) must be one.
Thus, the CDF \(H_n\) starts at 0 and ends at 1, as every valid distribution function should.
Final result:
\[ \boxed{H_n(t) = 1 - [1 - F(t)]^n} \]
From part (b), the cumulative distribution function of the sample minimum is
\[ H_n(t) = 1 - [1-F(t)]^n. \]
Since the \(X_i\) are continuous, the function \(F\) is differentiable with density \(f=F'\). The density of the sample minimum is therefore obtained by differentiating the CDF:
\[ h_n(t) := \frac{d}{dt} H_n(t). \]
Let \(u(t) = 1 - F(t)\), so that
\[
H_n(t) = 1 - [u(t)]^n.
\] Differentiating gives
\[
h_n(t) = -\frac{d}{dt}[u(t)]^n.
\]
By the chain rule,
\[
\frac{d}{dt}[u(t)]^n = n[u(t)]^{\,n-1} \cdot u'(t).
\]
Since \(u'(t) = \frac{d}{dt}(1-F(t)) =
-F'(t) = -f(t)\), it follows that
\[
h_n(t) = -n[1-F(t)]^{\,n-1}(-f(t)) = n[1-F(t)]^{\,n-1} f(t).
\]
Consistency with the support of the distribution
As in previous parts, the formula must agree with the natural behavior of the distribution at the boundaries. Let the support be \((a,b)\), where \(a\) or \(b\) may be infinite.
Case 1: \(t \leq
a\).
Outside the lower bound the distribution is constant, so \(f(t)=0\) and hence \(h_n(t)=0\).
This is expected: if \(t\) is smaller than the minimum possible value of the distribution, the density of the sample minimum must vanish.
Case 2: \(t \geq
b\).
Outside the upper bound the distribution is also constant, so again
\(f(t)=0\) and \(h_n(t)=0\).
This is consistent: if \(t\) is larger than the maximum possible value of the distribution, the density of the sample minimum must also vanish.
On the interval \((a,b)\), the density is nonnegative and integrates to one. Indeed, \[ \int_a^b h_n(t)\,dt = \int_a^b n[1-F(t)]^{n-1}f(t)\,dt. \] With the substitution \(w=1-F(t)\), one has \(dw=-f(t)\,dt\). When \(t=a\), it holds that \(F(a)=0 \Rightarrow w=1\), and when \(t=b\), it holds that \(F(b)=1 \Rightarrow w=0\). The integral becomes \[ n\int_1^0 w^{\,n-1}(-dw) = n\int_0^1 w^{\,n-1}\,dw = 1. \]
Thus, the density \(h_n\) is valid and consistent with the boundaries of the support.
Final result:
\[ \boxed{\,h_n(t) = n[1-F(t)]^{\,n-1} f(t)\ \text{for } t \in (a,b), \quad h_n(t)=0\ \text{otherwise.}\,} \]