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Ray Of Light [21]
3 years ago
7

A system has one each of two different types of components in joint operation. Let X and Y denote the random lengths of the live

s, in hundreds of hours, for components of type I and type II, respectively. The joint density function is:________

Mathematics
1 answer:
RSB [31]3 years ago
3 0

Answer:

The value of E(Y/X) is 2.

Step-by-step explanation:

As the complete question is not given, thus the complete question is found online and is attached herewith.

So the joint density function is given as

f(x,y)=\left \{ {{\dfrac{x}{8}e^{-\dfrac{x+y}{2}}    \,\,\,\,0 \leq x,0\leq y \atop {0}} \right.

So the marginal function for X is given as

f_x=\int\limits^{\infty}_0 {f(x,y)} \, dy\\f_x=\int\limits^{\infty}_0 \dfrac{x}{8}e^{-\frac{x+y}{2} }dy\\f_x=\int\limits^{\infty}_0 \dfrac{x}{8}e^{-\frac{x}{2}}e^{-\frac{y}{2} }dy\\f_x= \dfrac{x}{8}e^{-\frac{x}{2}}\int\limits^{\infty}_0e^{-\frac{y}{2} }dy\\f_x= \dfrac{x}{4}e^{-\frac{x}{2}}

Now

f(Y/X)=\dfrac{f(X,Y)}{f(X)}\\f(Y/X)=\dfrac{\dfrac{x}{8}e^{-\frac{x+y}{2} }}{\dfrac{x}{4}e^{-\frac{x}{2}}}\\f(Y/X)=\dfrac{1}{2}e^{-\frac{y}{2} }

Now the value of E(Y/X) is given as

E(Y/X)=\int\limits^{\infty}_0 {yf_{Y/X}} \, dy \\E(Y/X)=\int\limits^{\infty}_0 {y\dfrac{1}{2}e^{-\frac{y}{2} } }\, dy\\E(Y/X)=\dfrac{1}{2}\dfrac{\sqrt{2}}{(\dfrac{1}{2})^2}=2

So the value of E(Y/X) is 2.

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