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Question:-01

  1. 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)Vopt (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)Vprop (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)VRandom (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}}Vopt, V prop V prop V_(“prop”)V_{\text{prop}}Vprop, and V ran V ran V_(“ran”)V_{\text{ran}}Vran, 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)Vopt (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)Vprop (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)VRandom (x¯st ).

Justification

Given the provided equations and derivations, we have:
  1. For V ran − V prop V ran − V prop V_(“ran”)-V_(“prop”)V_{\text{ran}} – V_{\text{prop}}Vran−Vprop:
    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) >= 0V_{\text{ran}} – V_{\text{prop}} = \frac{\sum N_i\left(\overline{x_i}-\bar{x}\right)^2}{n N} \geq 0Vran−Vprop=∑Ni(xi¯−x¯)2nN≥0
    This implies that V ran ≥ V prop V ran ≥ V prop V_(“ran”) >= V_(“prop”)V_{\text{ran}} \geq V_{\text{prop}}Vran≥Vprop.
  2. For V prop − V opt V prop − V opt V_(“prop”)-V_(“opt”)V_{\text{prop}} – V_{\text{opt}}Vprop−Vopt:
    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)] >= 0V_{\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 0Vprop−Vopt=1nN[∑i=1kNi(Si−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}}Vprop≥Vopt.

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)Vopt(x¯st)≤Vprop(x¯st)≤Vran(x¯st)
This validates the given statement as False, with V opt V opt V_(“opt”)V_{\text{opt}}Vopt being the lower bound, V ran V ran V_(“ran”)V_{\text{ran}}Vran being the upper bound, and V prop V prop V_(“prop”)V_{\text{prop}}Vprop lying between them.
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