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drawar
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I'm practicing the past year papers to prepare for my coming finals. Please make necessary corrections if you feel something wrong with it, thanks!
Also, I'm supposed to do this in less than half an hour, so any suggestions on how to shorten this answer is really much appreciated!
The joint density function of X and Y is given by
f(x,y) = \frac{K}{y}{e^{ - \frac{{{y^2}}}{2} - \frac{x}{y}}} for 0 < y < x < \infty
and 0 elsewhere, where K is the normalizing constant.
(i) Determine the value of K.
(ii) Find the marginal probability density function of Y .
(iii) Evaluate E[Y] and Var(Y).
(iv) Find the conditional density function {f_{X|Y}}(x|y) for 0 < y < x, and then evaluate E[X|Y].
(v) Evaluate E[X].
(vi) Evaluate Cov(X; Y ).
(i)
\int\limits_0^\infty {\int\limits_0^x {\frac{K}{y}{e^{ - \frac{{{y^2}}}{2} - \frac{x}{y}}}dydx = } } K\int\limits_0^\infty {\int\limits_y^\infty {\frac{1}{y}{e^{ - \frac{{{y^2}}}{2} - \frac{x}{y}}}dxdy = } } K\int\limits_0^\infty {\frac{1}{y}{e^{ - \frac{{{y^2}}}{2}}}} \int\limits_y^\infty {{e^{ - \frac{x}{y}}}} dxdy = K\int\limits_0^\infty {\frac{1}{y}{e^{ - \frac{{{y^2}}}{2}}}y{e^{ - 1}}} dy = K{e^{ - 1}}\sqrt {2\pi } \frac{1}{{\sqrt {2\pi } }}\int\limits_0^\infty {{e^{ - \frac{{{y^2}}}{2}}}} dy = K{e^{ - 1}}\sqrt {2\pi } \frac{1}{2} because here I think y can be seen to follow a standard normal distribution whose graph is symmetrical about 0, so the area of the right half of the curve, which is equal to \frac{1}{{\sqrt {2\pi } }}\int\limits_0^\infty {{e^{ - \frac{{{y^2}}}{2}}}} dy is 1/2.
The above integral must equal 1, hence we solve K = e\sqrt {\frac{2}{\pi }}
(ii) It follows directly from (i) that {f_Y}(y) = K\int\limits_y^\infty {\frac{1}{y}{e^{ - \frac{{{y^2}}}{2} - \frac{x}{y}}}dx = K} {e^{ - \frac{{{y^2}}}{2}}}{e^{ - 1}} = {e^{ - \frac{{{y^2}}}{2}}}\sqrt {\frac{2}{\pi }}
(iii)E[Y] = \int\limits_0^\infty y {e^{ - \frac{{{y^2}}}{2}}}\sqrt {\frac{2}{\pi }} dy = \sqrt {\frac{2}{\pi }}
Let Z ~ N(0,1) => E[Z^2]=Var(Z)=1. Thus E[{Z^2}] = \frac{1}{{\sqrt {2\pi } }}\int\limits_{ - \infty }^\infty {{z^2}{e^{ - \frac{{{z^2}}}{2}}}dz = 1} \Rightarrow \frac{1}{{\sqrt {2\pi } }}\int\limits_0^\infty {{z^2}{e^{ - \frac{{{z^2}}}{2}}}dz = \frac{1}{2}}, and then
E[{Y^2}] = \sqrt {\frac{2}{\pi }} \int\limits_0^\infty {{y^2}} {e^{ - \frac{{{y^2}}}{2}}}dy = \sqrt {\frac{2}{\pi }} \int\limits_0^\infty {{z^2}{e^{ - \frac{{{z^2}}}{2}}}dz = } \sqrt {\frac{2}{\pi }} \times \frac{1}{2} \times \sqrt {2\pi } = 1
Var(Y) = E[{Y^2}] - E{[Y]^2} = 1 - \frac{2}{\pi } = \frac{{\pi - 2}}{\pi }
(iv)
{f_{X|Y}}(x|y) = \frac{{f(x,y)}}{{{f_Y}(y)}} = \frac{1}{y}{e^{1 - \frac{x}{y}}}
E[X|Y = y] = \frac{e}{y}\int\limits_0^\infty {x{e^{ - \frac{x}{y}}}} dx = \frac{e}{y}\left( {\frac{{{y^2} + y}}{e}} \right) = y + 1
(v)
E[X] = E[E[X|Y]] = E[Y + 1] = 1 + E[Y] = 1 + \sqrt {\frac{2}{\pi }}
(vi)
E[XY] = E[E[XY|Y]] = E[YE[X|Y]] = E[Y(Y + 1)] = E[{Y^2}] + E[Y] = 1 + \sqrt {\frac{2}{\pi }}
Cov(X,Y) = E[XY] - E[X]E[Y] = 1 + \sqrt {\frac{2}{\pi }} - \left( {1 + \sqrt {\frac{2}{\pi }} } \right)\sqrt {\frac{2}{\pi }} = 1 - \frac{2}{\pi }
Also, I'm supposed to do this in less than half an hour, so any suggestions on how to shorten this answer is really much appreciated!
Homework Statement
The joint density function of X and Y is given by
f(x,y) = \frac{K}{y}{e^{ - \frac{{{y^2}}}{2} - \frac{x}{y}}} for 0 < y < x < \infty
and 0 elsewhere, where K is the normalizing constant.
(i) Determine the value of K.
(ii) Find the marginal probability density function of Y .
(iii) Evaluate E[Y] and Var(Y).
(iv) Find the conditional density function {f_{X|Y}}(x|y) for 0 < y < x, and then evaluate E[X|Y].
(v) Evaluate E[X].
(vi) Evaluate Cov(X; Y ).
Homework Equations
The Attempt at a Solution
(i)
\int\limits_0^\infty {\int\limits_0^x {\frac{K}{y}{e^{ - \frac{{{y^2}}}{2} - \frac{x}{y}}}dydx = } } K\int\limits_0^\infty {\int\limits_y^\infty {\frac{1}{y}{e^{ - \frac{{{y^2}}}{2} - \frac{x}{y}}}dxdy = } } K\int\limits_0^\infty {\frac{1}{y}{e^{ - \frac{{{y^2}}}{2}}}} \int\limits_y^\infty {{e^{ - \frac{x}{y}}}} dxdy = K\int\limits_0^\infty {\frac{1}{y}{e^{ - \frac{{{y^2}}}{2}}}y{e^{ - 1}}} dy = K{e^{ - 1}}\sqrt {2\pi } \frac{1}{{\sqrt {2\pi } }}\int\limits_0^\infty {{e^{ - \frac{{{y^2}}}{2}}}} dy = K{e^{ - 1}}\sqrt {2\pi } \frac{1}{2} because here I think y can be seen to follow a standard normal distribution whose graph is symmetrical about 0, so the area of the right half of the curve, which is equal to \frac{1}{{\sqrt {2\pi } }}\int\limits_0^\infty {{e^{ - \frac{{{y^2}}}{2}}}} dy is 1/2.
The above integral must equal 1, hence we solve K = e\sqrt {\frac{2}{\pi }}
(ii) It follows directly from (i) that {f_Y}(y) = K\int\limits_y^\infty {\frac{1}{y}{e^{ - \frac{{{y^2}}}{2} - \frac{x}{y}}}dx = K} {e^{ - \frac{{{y^2}}}{2}}}{e^{ - 1}} = {e^{ - \frac{{{y^2}}}{2}}}\sqrt {\frac{2}{\pi }}
(iii)E[Y] = \int\limits_0^\infty y {e^{ - \frac{{{y^2}}}{2}}}\sqrt {\frac{2}{\pi }} dy = \sqrt {\frac{2}{\pi }}
Let Z ~ N(0,1) => E[Z^2]=Var(Z)=1. Thus E[{Z^2}] = \frac{1}{{\sqrt {2\pi } }}\int\limits_{ - \infty }^\infty {{z^2}{e^{ - \frac{{{z^2}}}{2}}}dz = 1} \Rightarrow \frac{1}{{\sqrt {2\pi } }}\int\limits_0^\infty {{z^2}{e^{ - \frac{{{z^2}}}{2}}}dz = \frac{1}{2}}, and then
E[{Y^2}] = \sqrt {\frac{2}{\pi }} \int\limits_0^\infty {{y^2}} {e^{ - \frac{{{y^2}}}{2}}}dy = \sqrt {\frac{2}{\pi }} \int\limits_0^\infty {{z^2}{e^{ - \frac{{{z^2}}}{2}}}dz = } \sqrt {\frac{2}{\pi }} \times \frac{1}{2} \times \sqrt {2\pi } = 1
Var(Y) = E[{Y^2}] - E{[Y]^2} = 1 - \frac{2}{\pi } = \frac{{\pi - 2}}{\pi }
(iv)
{f_{X|Y}}(x|y) = \frac{{f(x,y)}}{{{f_Y}(y)}} = \frac{1}{y}{e^{1 - \frac{x}{y}}}
E[X|Y = y] = \frac{e}{y}\int\limits_0^\infty {x{e^{ - \frac{x}{y}}}} dx = \frac{e}{y}\left( {\frac{{{y^2} + y}}{e}} \right) = y + 1
(v)
E[X] = E[E[X|Y]] = E[Y + 1] = 1 + E[Y] = 1 + \sqrt {\frac{2}{\pi }}
(vi)
E[XY] = E[E[XY|Y]] = E[YE[X|Y]] = E[Y(Y + 1)] = E[{Y^2}] + E[Y] = 1 + \sqrt {\frac{2}{\pi }}
Cov(X,Y) = E[XY] - E[X]E[Y] = 1 + \sqrt {\frac{2}{\pi }} - \left( {1 + \sqrt {\frac{2}{\pi }} } \right)\sqrt {\frac{2}{\pi }} = 1 - \frac{2}{\pi }