Let λ be the parameter of exponential distribution
So, for exponentially distributed random variables X & Y
Expected Value or Mean \(E\left( X \right) = \frac{1}{\lambda }\)
Variance: \(Var\;\left( X \right) = \frac{1}{{{\lambda ^2}}}\)
Var (X + Y) = Var X + Var Y + 2 Cov (X, Y)
Where Cov (X, Y) = Covariance of X and Y
Correction coefficient is given by:
\(r = \frac{{Cov\;\left( {X,\;Y} \right)}}{{\sqrt {Var\left( X \right)Var\left( Y \right)} }}\)
Calculation:
Given: E(X) = E(Y) = 0.5
\(\frac{1}{{{\lambda _1}}} = \frac{1}{{{\lambda _2}}} = 0.5\)
λ_{1} = λ_{2} = 2
\(Var\;\left( X \right) = \frac{1}{{\lambda _1^2}} = \frac{1}{{{2^2}}} = 0.25\)
\(Var\;\left( Y \right) = \frac{1}{{\lambda _2^2}} = \frac{1}{{{2^2}}} = 0.25\)
Also it is given, Z = X + Y
And Var (Z) = 0
∴ Var (X + Y) = 0
Var X + Var Y + 2 Cov (X, Y) = 0
On putting values:
0.25 + 0.25 + 2 Cov (X, Y) = 0
\(Cov\;\left( {X,\;Y} \right) =  \frac{{0.5}}{2} =  0.25\)
Correction coefficient is given by:
\(r = \frac{{Cov\;\left( {X,\;Y} \right)}}{{\sqrt {Var\left( X \right)Var\left( Y \right)} }} = \frac{{  0.25}}{{\sqrt {0.25 \times 0.25} }} = \;  1\)Given
r12 = r13 = r23 = r
Formula
r_{12.3} = (r_{12}  r_{13}.r_{23})/√[(1  r_{12}^{2})(1  r_{23}^{2})
Calculation
According to question
⇒ (r  r × r)/√(1  r^{2})(1  r^{2})
⇒ (r  r^{2})/(1  r^{2})
⇒ r(1  r)/(1  r)(1 + r)
⇒ The value of r_{12.3} is r/(1 + r)
Given
The correlation coefficient between two variables X and Y = 0.4
Concept used
The correlation coefficient (r) is independent of origin and scale and depend on the sign of variables
Calculation
The correlation coefficient between the two variables is the measure of the slope between the variables in the regression graph. It is given that the correlation coefficient between X and Y is 0.4 and the correlation coefficient is independent of change of origin and scale but it depends on variables
∴ The correlation coefficient between 2X and (Y) is  0.4
The value of simple correlation coefficient in the interval of [1, 1]
The regression coefficient is independent of the change of origin. But, they are not independent of the change of the scale. It means there will be no effect on the regression coefficient if any constant is subtracted from the values of x and y
Given
Multiple correlation coefficient = R_{1.23}
Explanation
For multiple correlation coefficient,
⇒ R_{1.23} ≥ r_{12} × r_{13} × r_{23}
From given expression we can say that
∴ The multiple correlation coefficient R_{1,23} as compared to any simple correlation coefficients between the distinct variable X_{1} ,X_{2}, and X_{3} is not less than any r_{12}, r_{13} and r_{23}
Given
b_{xy} = 0.5
b_{yx} = 405
Formula
The correlation coefficient is the geometric mean of two regression coefficient
r = √(b_{xy} × byx)
r = correlation coefficient
Calculation
r = √(0.5 × 4.5)
r = 1.5
∴ Correlation coefficient is 1.5
Explanation:
The general formula for multiple correlation coefficient
\(\rm R^2_1{.}{_2}{_3} = \frac{r_{12}^{2}+r_{13}^{2}2r_{12}r_{13}r_{23}}{1r_{23}^{2}}\)
A multiple correlation coefficient is a nonnegative coefficient.
It is the value ranges between 0 and 1. It cannot assume a minus value.
If the multiple correlation coefficient of X1 on X2 and X3 is zero, then:
If R1.23 = 0 then r12 = 0 and r13 = 0
R1.23 is the same as R1.32
Determine the correlation coefficient between the pulses x(t) and y(t) shown in the fig. below:
Cross correlation between x(t) and y(t) is:
C_{o}(t) = x(t) * y(t)
\( = \mathop \smallint \nolimits_{  \infty }^\infty x\left( \tau \right)y\left( {t + \tau } \right)d\tau \)
x(t) = 1 0 ≤ τ ≤ 2
y(t) = sin 2πt
\(C_0 \left( t \right) = \mathop \smallint \nolimits_0^2 1 \cdot \sin \left( {2\pi \left( {t + \tau } \right)} \right)d\tau = \mathop \smallint \nolimits_0^2 \sin 2\;\pi t \cdot \cos 2\pi \;\tau d\tau + \mathop \smallint \nolimits_0^2 \cos 2\pi t \cdot \sin 2\pi t\)
\( \Rightarrow \sin 2\pi t\;\mathop \smallint \nolimits_0^2 \cos 2\pi \tau d\tau + \cos 2\pi t\;\mathop \smallint \nolimits_0^2 \sin 2t\)
\( \Rightarrow \frac{{\sin 2\pi t}}{{2\pi }}[\sin 2\pi \tau ]_0^2 + \frac{{\cos 2\pi t}}{{2\pi }}\left[ {  \cos 2\pi \tau } \right]_0^2\)
⇒ 0
Important Points
Whenever we integrate sine or cosine in its one time period (T) or integer multiple of its time period (nT) then its integration will be zero.
Concept:
The Fourier transform of R_{xx}(τ) in defined as:
\({R_{xx}}\left( \tau \right)\mathop \leftrightarrow \limits^{\;\;F.T.\;\;} {S_{xx}}\left( \omega \right)\)
S_{xx}(ω) = Energy Spectral Density defined as:
\({S_{xx}}\left( \omega \right) = \;\mathop \smallint \limits_{  \infty }^\infty {R_{xx}}\left( \tau \right){e^{  j\omega \tau }}d\omega \)
The term on the righthand side is the Fourier transform of R_{xx}(τ)Formula
If θ be the angle between the regression lines then,
Tan θ = [(1  r^{2})/r] × (σ_{x} × σ_{y})/(σ_{x}^{2} + σ_{y}^{2})
Explanation
According to question
⇒ σ_{x} = σ_{y}
⇒ Tanθ = [(1  r^{2}/r] × (σ_{x}^{2}/(2σ_{x})
⇒ Tanθ = [(1  r^{2}/r] × 1/2
⇒ Tan θ = (1  r^{2}/2r
⇒ Tanθ = Sinθ/cosθ
⇒ Tanθ = Sinθ × cosθ
⇒ Sinθ = Tanθ /secθ
⇒ Sinθ = Tanθ/√(1 + tan^{2}θ)
⇒ Sinθ = [(1  r^{2}/r]/√(1 + (1  r^{2}/2r)^{2})
⇒ Sinθ = (1  r^{2})/√[(1  r^{2})^{2} + 4r^{2}]
∴ \(sin \theta=\sqrt{\dfrac{1r^2_{xy}}{1+r^2_{xy}}}\)
Explanation
The simplest device for ascertaining whether two variables are related is to prepare a dot chart called scatter diagram. When this method is used the given data are plotted on a graph paper in the form of dots.
∴ If a data set contains n paired values on two variables x(independent) and y(dependent), then their plot is called Scatter diagram
Calculation
The regression line of Y on X is Y = 30  0.9x
⇒ Y  30 =  0.9x
The regression equation line of Y on X is = y  y_{1} = r(s_{y}/s_{x})(x  x_{1})
Comparing both equations, we get
⇒ r(s_{y}/s_{x}) = 0.9
⇒ r(9/2) = 0.9
⇒ r = (0/9 × 2)/9 =  0.2
∴ The value of the correlation coefficient rxy is 0.2
Given
σ_{x} = √16 = 4
r = 0.48
Covariance = ∑xy/N = 36
Fomula
Covariance = ∑xy/N
r = ∑xy/N.σ_{x} × σ_{y}
Calculation
According to question
⇒ 0.48 = 36/4 × σ_{y}
⇒ σ_{y} = 9/0.48
∴ The standard deviation of y(σ_{y}) is 18.75
Given
Let θ be the angle made by the line of regression of Y on X.
σY = 2σX and the correlation coefficient between X and Y = γ = 0.3
Calculation
Let L_{YX} be line of regression of Y on X
Let the angle made by L_{yx} with x – axis be θ.
∴ Tangent of the angle made by L_{yx} with x – axis is tan θ.
But slope of any line with x – axis is tan θ
⇒ Slope of L_{yx} = b_{yx}
⇒ b_{yx} = tan θ
⇒ b_{yx} = γ(σ_{y}/σ_{x})
⇒ tan θ = 0.3 × 2(σ_{x}/σ_{x})
⇒ tan θ = 0.3 × 2
∴ The value of θ is tan^{1}(0.6)
Explanation
Let each observation be y_{xi}
Where x indicates the category that observation is in and
i is the label of the particular observation.
n_{x} = The number of observation in category x
⇒ y̅_{x}= ∑iy_{xi}/n_{x}
⇒ y̅ = ∑_{x}n_{x}y̅_{x}/∑_{x}n_{x}
y̅_{x} = Mean of category x
y̅ = mean of the whole population
The correlation ratio E^{2} = ∑_{x}n_{x}(y̅_{x} – y̅)^{2}/∑_{xi}(y_{xi} – y̅)^{2}
⇒ E^{2} = σ^{2}_{y̅ }/σ^{2}_{y}
Where as, σ^{2}_{y̅ } = ∑_{x}n_{x}(y̅_{x} – y̅)^{2}/∑_{x}n_{x}
σ^{2}_{y} = ∑_{x,i}(y_{xi} – y̅)^{2}/n
The weighted variance of the category means divided by the variance of all samples.It is the relation between values of x and values of y̅_{x} is linear this will give the same result as the square of pearson’s correlation coefficient; otherwise the correlation ratio will be larger in magnitude. It can therefore be used for judging non linear relationship
∴ E takes the values between 0 and 1
Important Points
The E = 0 represents the special case of no dispersion among the mean of the different categories
E = 1 refers to no dispersion within the respective categories
E^{2} = 0 to 1, we have to learn this result
The value of correlation coefficient =  1 to 1
Explanation
The correlation Coefficient is a statistical measure of the strength of the relationship between the relative movements of two variables. The value range between 1 to 1
A calculated number greater than 1 or less than 1 means thta there was an error in the correlation measurement
Correlation 1 means a perfect positive correlation
Correlation 1 means a perfect negative correlation
0 correlation shows no linear relationship between the movement of the two variables
∴ The value of simple correlation coefficient in the interval of [1, 1]
Explanation
The coefficient of regression is given as = β_{xIy} and β_{γIx}
We know that the coefficient of correlation r = ± √(b_{xy}_{ }× b_{yx}) where b_{xy} and b_{yx} are coefficient of regression
⇒ Similarly from the above options there is only 1 option that satisfied this equation
∴ The coefficient of correlation is \(\pm\sqrt{{\beta_{Xy}\beta_{Yx}}}\)
We are given that X, Y, Z are uncorrelated and have variances \(σ_x^2, σ_y^2 \:and\:σ_z^2\) respectively,
Then by the definition of the correlation coefficient
\(ρ_{X + Y, Y + Z} =\dfrac{cov(X + Y, Y + Z)}{\sqrt{Var(X+Y)\cdot Var(Y+Z)}} \)
Lets compute the numerator and denominator of the above expression individually
Numerator:
cov(X + Y, Y + Z) = cov(X, Y) + cov(X, Z) + cov(Y, Y) + cov(Y, Z)
cov(X + Y, Y + Z) = 0 + 0 + σ_{y}^{2} + 0
cov(X + Y, Y + Z) = σy2
Denominator :
Var(X + Y) = Var(X) + Var(Y) (Given X and Y are uncorrelated)
Var(X + Y) = σx2 + σy2
Var(Y + Z) = Var(Y) + Var(Z) (Given Y and Z are uncorrelated)
Var(Y + Z) = σy2 + σ_{z}2
Finally, we have the following expression for the correlation coefficient
\(ρ_{X + Y, Y + Z} =\dfrac{σ_y^2}{\sqrt{(σ_x^2 + σ_y^2)\cdot (σ_y^2 + σ_z^2)}} \)
Hence correct option will be none of these.
Note:
If given σx2 = σy2 = σz2 = σ^{2 } (Equal variences) then
Correlation coefficient is given as,
\(ρ_{X + Y, Y + Z} =\dfrac{σ^2 }{\sqrt{(2σ^2)\cdot (2σ^2)}} = \dfrac12\)
Calculate the correlation coefficient between the following values :
x: 3, 5, 1, 7, 5
y: 4, 3, 0, 8, 2
Given
x 
3 
5 
1 
7 
5 
y 
4 
3 
0 
8 
2 
Formula
Correlation coefficient = n∑xy – (∑x × ∑ y)/√{[n∑x^{2} – (∑x)^{2}] × [n∑y^{2} – (∑y)^{2}}
n∑xy = arithmetic mean of the summition of product of x and y
x and y = arithmetic average of x and y series
n∑x^{2} and n∑y^{2} = arithmetic mean of sum of sauare of item of x and y
Calculation

x 
y 
x^{2} 
y^{2} 
Xy 

3 
4 
9 
16 
12 

5 
3 
25 
9 
15 

1 
0 
1 
0 
0 

7 
8 
49 
64 
56 

5 
2 
25 
4 
10 
Total 
21 
17 
109 
93 
93 
Putting these value in formula
⇒ r = 5 × 93 – 21 × 17/√{[5 × 109 – (21)^{2}] × [5 × 93  (17)^{2}}
⇒ r = 108/√[104 × 176]
⇒ r = 108/135
⇒ r = 0.8
∴ The correlation coefficient between x and y are 0.8
Explaination
Remember the result that the range of values that coefficient of correlation can take are 1 ≤ r ≤ +1
Important Points
Basically, Correlation refers to the relationship of variables . There are 3 types of correlation
1  Positive or negative correlation
2  Simple, partial or multiple
3  Linear or non linear
Formula
The general formula of multiple correlation coefficient
\(\rm R^2_1{.}{_2}{_3} = \frac{r_{12}^{2}+r_{13}^{2}2r_{12}r_{13}r_{23}}{1r_{23}^{2}}\)
Explanation:
Multiple correlation coefficient is a nonnegative coefficient.
It is the value ranges between 0 and 1. It cannot assume a minus value.
If R_{1.23} = 0 then r_{12} = 0 and r_{13} = 0
R_{1.23} is the same as R_{1.32}