Question:-01
State whether the following statements are true or false and also give the reason in support of your answer:
(a)
V
opt
(
x
¯
st
)
V
opt
x
¯
st
V_(“opt “)( bar(x)_(“st “)) V_{\text {opt }}\left(\bar{x}_{\text {st }}\right) V opt ( x ¯ st ) lies between
V
prop
(
x
¯
st
)
V
prop
x
¯
st
V_(“prop “)( bar(x)_(“st “)) V_{\text {prop }}\left(\bar{x}_{\text {st }}\right) V prop ( x ¯ st ) and
V
Random
(
x
¯
st
)
V
Random
x
¯
st
V_(“Random “)( bar(x)_(“st “)) V_{\text {Random }}\left(\bar{x}_{\text {st }}\right) V Random ( x ¯ st ) .
Answer:
Introduction
We are given three different expressions representing variances or measures of dispersion, namely
V
opt
V
opt
V_(“opt”) V_{\text{opt}} V opt ,
V
prop
V
prop
V_(“prop”) V_{\text{prop}} V prop , and
V
ran
V
ran
V_(“ran”) V_{\text{ran}} V ran , and we are asked to verify the statement:
V
opt
(
x
¯
st
)
V
opt
x
¯
st
V_(“opt “)( bar(x)_(“st “)) V_{\text {opt }}\left(\bar{x}_{\text {st }}\right) V opt ( x ¯ st ) lies between
V
prop
(
x
¯
st
)
V
prop
x
¯
st
V_(“prop “)( bar(x)_(“st “)) V_{\text {prop }}\left(\bar{x}_{\text {st }}\right) V prop ( x ¯ st ) and
V
Random
(
x
¯
st
)
V
Random
x
¯
st
V_(“Random “)( bar(x)_(“st “)) V_{\text {Random }}\left(\bar{x}_{\text {st }}\right) V Random ( x ¯ st ) .
Justification
Given the provided equations and derivations, we have:
For
V
ran
−
V
prop
V
ran
−
V
prop
V_(“ran”)-V_(“prop”) V_{\text{ran}} – V_{\text{prop}} V ran − V prop :
V
ran
−
V
prop
=
∑
N
i
(
x
i
¯
−
x
¯
)
2
n
N
≥
0
V
ran
−
V
prop
=
∑
N
i
x
i
¯
−
x
¯
2
n
N
≥
0
V_(“ran”)-V_(“prop”)=(sumN_(i)( bar(x_(i))-( bar(x)))^(2))/(nN) >= 0 V_{\text{ran}} – V_{\text{prop}} = \frac{\sum N_i\left(\overline{x_i}-\bar{x}\right)^2}{n N} \geq 0 V ran − V prop = ∑ N i ( x i ¯ − x ¯ ) 2 n N ≥ 0
This implies that
V
ran
≥
V
prop
V
ran
≥
V
prop
V_(“ran”) >= V_(“prop”) V_{\text{ran}} \geq V_{\text{prop}} V ran ≥ V prop .
For
V
prop
−
V
opt
V
prop
−
V
opt
V_(“prop”)-V_(“opt”) V_{\text{prop}} – V_{\text{opt}} V prop − V opt :
V
prop
−
V
opt
=
1
n
N
[
∑
i
=
1
k
N
i
(
S
i
−
S
¯
)
2
]
≥
0
V
prop
−
V
opt
=
1
n
N
∑
i
=
1
k
N
i
S
i
−
S
¯
2
≥
0
V_(“prop”)-V_(“opt”)=(1)/(nN)[sum_(i=1)^(k)N_(i)(S_(i)-( bar(S)))^(2)] >= 0 V_{\text{prop}} – V_{\text{opt}} = \frac{1}{n N}\left[\sum_{i=1}^k N_i\left(S_i-\bar{S}\right)^2\right] \geq 0 V prop − V opt = 1 n N [ ∑ i = 1 k N i ( S i − S ¯ ) 2 ] ≥ 0
This implies that
V
prop
≥
V
opt
V
prop
≥
V
opt
V_(“prop”) >= V_(“opt”) V_{\text{prop}} \geq V_{\text{opt}} V prop ≥ V opt .
Conclusion
Combining the above two results, we can conclude that:
V
opt
(
x
¯
st
)
≤
V
prop
(
x
¯
st
)
≤
V
ran
(
x
¯
st
)
V
opt
x
¯
st
≤
V
prop
x
¯
st
≤
V
ran
x
¯
st
V_(“opt”)( bar(x)_(“st”)) <= V_(“prop”)( bar(x)_(“st”)) <= V_(“ran”)( bar(x)_(“st”)) V_{\text{opt}}\left(\bar{x}_{\text{st}}\right) \leq V_{\text{prop}}\left(\bar{x}_{\text{st}}\right) \leq V_{\text{ran}}\left(\bar{x}_{\text{st}}\right) V opt ( x ¯ st ) ≤ V prop ( x ¯ st ) ≤ V ran ( x ¯ st )
This validates the given statement as False, with
V
opt
V
opt
V_(“opt”) V_{\text{opt}} V opt being the lower bound,
V
ran
V
ran
V_(“ran”) V_{\text{ran}} V ran being the upper bound, and
V
prop
V
prop
V_(“prop”) V_{\text{prop}} V prop lying between them.