Question

Let $Z_1,Z_2......Z_n$be independent standard normal random variables, and let $S_j=\sum _{i=1}^j{Z}_i$

(a) What is the conditional distribution of Sn given that $S_k=y,$for k = 1, ... , n?

(b) Show that, for 1 … k … n, the conditional distribution of $S_K$given that

Sn = x is normal with mean xk/n and variance k(n − k)/n.

Short answer

a) The normal random variable $S_n$has a mean of zero and a variance of

n-k, which is independent of the $S_k$.

b) The conditional distribution of the function$S_{k}for1\le K\le n$

Step-by-step solution

Part (a)- Step:1 To find

The conditional distribution of$S_n$

Part (a) - Step 2: Explanation

We know that$S_J=\sum _{i=1}^J{Z}_i$

Note that$S_n-S_k=\sum _{i=k+1}^n{Z}_i$is a normal random variable with mean 0 and variance$\mathrm{n}-\mathrm{k}$that is independent of ${\mathrm{S}}_{\mathrm{k}}$, given that $S_k=y$is a normal random variable with mean $\mathrm{y}$and variance $n-k$

Part (b) Step 3:To find

The conditional distribution of$S_{K}for1\le K\le n$

Part (b) - Step 4: Explanation

Given : ${\mathrm{Z}}_1,{\mathrm{Z}}_2,\dots ..,{\mathrm{Z}}_{\mathrm{n}}$be independent standard normal random variables, and let

$S_j=\sum _{i=1}^j{Z}_i$

${\mathrm{S}}_{\mathrm{k}}=\mathrm{y}\text{for}\mathrm{k}=1,\dots .,\mathrm{n}-1\text{and}{\mathrm{s}}_{\mathrm{n}}=\mathrm{x}\text{with mean}\mathrm{xk}/\mathrm{n}\text{and variance}\mathrm{k}(\mathrm{n}-\mathrm{k})/\mathrm{n}.$

Calculation :Because the conditional density function of $S_K$given that

$S_n=X$is a density function with the argument Y, anything that is independent of y can be considered a constant. (For example, x is considered a fixed constant.) The values C i, i=1,2,3,4 in the following are all constants that are independent of Y.

$\begin{array}{r}{f}_{{\left.s_1\right|}_2}(y\mid x)=\frac{f_{s_1\mid s_0}(y,x)}{f_{s_0}\left(x\right)}\\ =C_1{f}_{s_0{l}_2}(x\mid y)f_{s_3}\left(y\right)\left(\text{where}{C}_1=\frac{1}{f_{s_e}\left(x\right)}\right)\\ =C_1\frac{1}{\sqrt{2\pi }\sqrt{n-k}}{e}^{-(x-y{)}^2/2(n-k)}\frac{1}{\sqrt{2\pi }\sqrt{k}}{e}^{-y^2/2k}\\ =C_2\exp \left\{\frac{-(x-y{)}^2}{2(n-k)}-\frac{y^2}{2k}\right\}\left(\because where{c}_2=\frac{c_1}{2\pi \sqrt{n-k}\sqrt{k}}\right)\\ =C_3\exp \left\{\frac{2xy}{2(n-k)}-\frac{y^2}{2(n-k)}-\frac{y^2}{2k}\right\}\left(\because where{c}_3=c_2\exp \left(\frac{-x^2}{2(n-k)}\right)\right)\\ =C_3\exp \left\{\frac{2xy}{2(n-k)}-\frac{y^2}{2}\left(\frac{1}{n-k}-\frac{1}{k}\right)\right\}\\ =C_3\exp \left\{\frac{2xy}{2(n-k)}-\frac{y^2}{2}\left(\frac{k+n-k}{k(n-k)}\right)\right\}\end{array}$

$\begin{array}{r}=C_3\exp \left\{\frac{2xy}{2(n-k)}-\frac{y^{2}n}{2k(n-k)}\right\}\\ =C_3\exp \left\{\frac{-n}{2k(n-k)}\left(y^2-\frac{2k}{n}xy\right)\right\}\\ =C_3\exp \left\{\frac{-n}{2k(n-k)}\left[{\left(y-\frac{k}{n}x\right)}^2-{\left(\frac{kx}{n}\right)}^2\right]\right\}\\ =C_4\exp \left\{\frac{-n}{2k(n-k)}{\left(y-\frac{k}{n}x\right)}^2\right\}\left(\because \text{where}{c}_4=c_3\exp \left\{-{\left(\frac{kx}{n}\right)}^2\right\}\right)\end{array}$

But we recognise the preceding as the density function of a normal random variable with mean $\frac{Kx}{n}$and variance $\frac{k(n-k)}{n}$