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If X and Y are two independent random variables such that X ~ χ2m and Y~ χ2 n , then the distribution of (X +Y) is
- a)normal.
- b)standard normal.
- c)T.
- d)chi-square.
Correct answer is option 'D'. Can you explain this answer?
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If X and Y are two independent random variables such that X ~ χ2m...
Distribution of (X Y) when X and Y are independent random variables
When two random variables X and Y are independent, the joint probability distribution of (X, Y) is the product of their marginal probability distributions.
Thus, if X ~ 2m and Y ~ 2n, then the joint probability distribution of (X, Y) is:
f(X, Y) = f(X) * f(Y)
where f(X) and f(Y) are the probability density functions of X and Y, respectively.
Finding the distribution of (X Y)
To find the distribution of (X Y), we need to find the probability density function of Z = X Y.
Let F(Z) be the cumulative distribution function of Z. Then:
F(Z) = P(Z ≤ z) = P(X Y ≤ z)
Since X and Y are independent, we can express the joint probability distribution of (X, Y) as:
f(X, Y) = f(X) * f(Y)
Thus, we can rewrite the probability density function of Z as:
f(Z) = ∫∫ f(X, Y) * δ(X Y - z) dX dY
where δ is the Dirac delta function.
Solving the integral, we get:
f(Z) = ∫∫ f(X) * f(Y) * δ(X Y - z) dX dY
= ∫∞-∞ f(X) * f(z/X) / |X| dX
where |X| denotes the absolute value of X.
This is the probability density function of Z, which is a chi-square distribution with 2 degrees of freedom.
Therefore, the distribution of (X Y) is a chi-square distribution with 2 degrees of freedom when X and Y are independent random variables with X ~ 2m and Y ~ 2n.
When two random variables X and Y are independent, the joint probability distribution of (X, Y) is the product of their marginal probability distributions.
Thus, if X ~ 2m and Y ~ 2n, then the joint probability distribution of (X, Y) is:
f(X, Y) = f(X) * f(Y)
where f(X) and f(Y) are the probability density functions of X and Y, respectively.
Finding the distribution of (X Y)
To find the distribution of (X Y), we need to find the probability density function of Z = X Y.
Let F(Z) be the cumulative distribution function of Z. Then:
F(Z) = P(Z ≤ z) = P(X Y ≤ z)
Since X and Y are independent, we can express the joint probability distribution of (X, Y) as:
f(X, Y) = f(X) * f(Y)
Thus, we can rewrite the probability density function of Z as:
f(Z) = ∫∫ f(X, Y) * δ(X Y - z) dX dY
where δ is the Dirac delta function.
Solving the integral, we get:
f(Z) = ∫∫ f(X) * f(Y) * δ(X Y - z) dX dY
= ∫∞-∞ f(X) * f(z/X) / |X| dX
where |X| denotes the absolute value of X.
This is the probability density function of Z, which is a chi-square distribution with 2 degrees of freedom.
Therefore, the distribution of (X Y) is a chi-square distribution with 2 degrees of freedom when X and Y are independent random variables with X ~ 2m and Y ~ 2n.
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D
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If X and Y are two independent random variables such that X ~ χ2m and Y~ χ2 n , then the distribution of (X +Y) isa)normal.b)standard normal.c)T.d)chi-square.Correct answer is option 'D'. Can you explain this answer?
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