Question

X and Y are two independent random variables with variances 1 and 2 respectively. Let Z=X-Y. The variance of Z is _____.

Answer 1

$Z = X - Y$

For independent random variables X and Y, the variance of their sum or difference is the sum of their variances

$\sigma^2_Z = \sigma^2_X  + \sigma^2_Y$

$\sigma^2_Z = 1 + 2  = 3$

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 what "Z = X - Y" means?

let X be a random variable for a six sided die 

$X = \{1,2,3,4,5,6 \}$

$E[X^2] = \large \frac{1}{6}$ $(1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2) = 15.166$

$E[X] = \large \frac{1}{6}$ $(1 + 2 + 3 + 4 + 5+ 6) = 3.5$

$(E[X])^2 = 3.5^2 = 12.25$

$\sigma^2_X =  E[X^2] -(E[X])^2   = 15.166 - 12.25$

$\sigma^2_X = 2.916$

let Y be a random variable of a four sided die

$Y = \{1,2,3,4\}$

$\sigma^2_Y = 1.25$

All we know about X and Y are the variables which results in values which are in specific domain.

Both are independent random variables as we are throwing both die simultaneously so result of 1 die doesn't affect the result of other.

Now we introduce a dependent random variable $Z$ = X - Y

Z will depend on values of X and Y and will be equal to X - Y.

$Z =\{-3,-2,-1,0,1,2,3,4,5 \}$

$P(-3) = P(5) = \frac{1}{24}$

$P(-2) = P(4) = \frac{2}{24}$

$P(-1) = P(3) = \frac{3}{24}$

$P(2) = P(1) = P(0) = \frac{4}{24}$

$E[Z] $= 1

$E[Z^2]$ = $\frac{1}{24}\left(\left(-3\right)^{2}+5^{2}\right) +\frac{2}{24}\left(\left(-2\right)^{2}+4^{2}\right)+\frac{3}{24}\left(\left(-1\right)^{2}+3^{2}\right)+\frac{4}{24}\left(0^{2}+1^{2}+2^{2}\right)  $

$ \large =\frac{34}{24}+\frac{40}{24}+\frac{30}{24}+\frac{20}{24}  =\frac{124}{24}=\frac{31}{6} $$= 5.166$

$\sigma^2_Z = 5.166 - 1 = \color{blue}{4.166}$

Now let's find out if we are getting same answer by applying the identity.

$\sigma^2_Z = \sigma^2_X  + \sigma^2_Y$

$\sigma^2_Z = 2.916 + 1.25 =  \color{blue}{4.166}$

This was a not a proof but an insight on how independent random variables affect variances on subtraction. Now calculate variance of $Z'$ where $Z' = X + Y$. You'll get the answer $\color{blue}{4.166}$

Comments

if question ask Z=X+Y,then answer remains same or not????

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remains the same

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why they are given $Z=X-Y$

What they mean by this?

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srestha explained in detail 

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Mk Utkarsh nice approach

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thanks bro :)

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srestha explained in more detail 

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Given that  $X$ and $Y$ are two independent random variables,$Variance (X)=1$ and $Variance (Y)=2$

                    $\sigma ^{2}_{X}=1,\sigma ^{2}_{Y}=2$

                         $Z=X-Y$

             $Variance(Z)=Variance(X)+Variance(-Y)$

            $Variance (-Y)=(-1)^{2}.Variance(Y)$         

$-->$ Variance is always $\geq 0$ because standard deviation$(SD)=\sqrt{\sigma^{2}}$ ,if $\sigma^{2}$ is negative then standard deviation is complex,which is not possible.

$Variance(Z)=Variance(X)+(-1)^{2}.Variance(Y)$

$Variance(Z)=Variance(X)+Variance(Y)$

            $\sigma ^{2}_{Z}=\sigma ^{2}_{X}+\sigma ^{2}_{Y}$

             $\sigma ^{2}_{Z}=1+2=3$

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@ srestha ma'am

$\sigma _{x}^{2}=\sqrt{1^{2}+2^{2}+3^{2}+4^{2}+5^{2}+6^{2}}=9.53$

How you write this equation?

If $X$ is a discrete random variable then 

$Mean$ $(or)$ $average$ $value$ $(or)$  $expected$ $value$ $(or)$ $expectation=\mu_{X}=E[X]=X_{i}P[X_{i}]$ $where$ $i=0,1,2,...$

$Variance=Var(X)=\sigma^{2}_{X}=E[X^{2}]-E[X]$

Here $E[X^{2}]=X^{2}_{i}P[X_{i}]$

$Standard$ $deviation(SD)=\sqrt{\sigma^{2}_{X}}=\sigma_{X}$

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yes, sorry

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Please, you do not say, sorry.

You are great.

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:p

ok, good night :)

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@Mk Utkarsh

in ur example, how u got value of $Z=-3??$

$Z=X-Y$ ,right??

X starts with $1$ and $Y$ too starts with $1$, right??

@Lakshman Patel RJIT

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@srestha ma'am

please see my above comment.

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@Lakshman Patel RJIT

yes, I got it now.

He just done, total number of distinct differences when

$X=\left \{ 1,2,3,4,5,6 \right \}$

and

$Y=\left \{ 1,2,3,4 \right \}$

right?

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Another thing tell me $P\left ( -3 \right )=P\left ( 5 \right )=\frac{1}{24}?$

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@srestha

Ma'am, I didn't understand the after this

$Z=\{ -3,-2,-1,0,1,2,3,4,5\}$

Where he take $Z$?

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yes, got it mostly

$X=\left \{ 1,2,3,4,5,6 \right \}$

$Y=\left \{ 1,2,3,4 \right \}$

Now, put $Z=X-Y$

and check what values u get?

Now as, $X$ has $6$ elements and $Y$ has $4$ elements, total $6\times 4=24$ elements

And $P(1-4)=P(3)$

and $P(6-1)=P(5)$ comes one time.

While P(0) comes $4$ times $P(1-1)=P(2-2)=P(3-3)=P(4-4)$ like this he gone

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I subtract the value $Z=X-Y$ and get only $Z=\{5,6\}$

i didn't understand where $Z=\{-3-2-1,0,1,2,3,4,5\}$ come?

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What is X and what is Y??

Have u subtracted every term of X from every term of Y?

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I subtract like set theory.

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                                                                      $Y$ 

     $X-Y$             $\overbrace{.............................................................}$

X	1	2	3	4
1	0	-1	-2	-3
2	1	0	-1	-2
3	2	1	0	-1
4	3	2	1	0
5	4	3	2	1
6	5	4	3	2

Z={−3−2−1,0,1,2,3,4,5 }. Select Distinct from Table.

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@Lakshman Patel RJIT

plz try to give detail answer, when u getting doubt

Now clear?

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@sresthaYes sure

$X=\{1,2,3,4,5,6\}$ and $Y=\{1,2,3,4\}$

$Z=X-Y$

$Z=\{1,2,3,4,5,6\}-\{1,2,3,4\} = \{5,6\}$

Why we subtract indivisual number?

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no.

Here each digit subtracting from each digit.

Take a digit of set X, say $1$

Now subtract it from each digit of Y={1,2,3,4}

What we get from it? {0,-1,-2,-3}

like this do for other digit of X

@Satbir done correctly. Check now

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yes thank you

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Sorry for being late at the party but any more doubts on this answer?

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@Mk Utkarsh

Chk this, According to you what should be the ans? According to this answer, option D) there also should be correct. right?

@Shaik Masthan Among C) and D) what should be ans?

 

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@srestha  option “D”  is the correct answer for this question because if 1 and X are independent then the Var(1,x) will be 3 and if they are dependent then Var(1,x) will be less than 3. Hence this is undecidable so option D is correct.