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Big-Daddy
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I saw a question where f(x) = (1/2)x was given as the probability density function for random variable X. Then they said variable Y follows the probability density function y=f(x-1) and asked to calculate E(Y) and Var(Y).

I tried to just use the E(Y) = E(aX+b) = a*E(X) + b theorem which would then give E(Y) = E(X)-1 for this case, but for some reason that's wrong and the answer is E(Y) = E(X) + 1 (and you can calculate E(X) from f(x) integrated of course). Why is this?

Edit: I can visualize why the mean would be shifted when you translate by one unit in the x-direction. But why does the calculation not work?
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ghostwalker
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(Original post by Big-Daddy)
I saw a question where f(x) = (1/2)x was given as the probability density function for random variable X. Then they said variable Y follows the probability density function y=f(x-1) and asked to calculate E(Y) and Var(Y).

I tried to just use the E(Y) = E(aX+b) = a*E(X) + b theorem which would then give E(Y) = E(X)-1 for this case, but for some reason that's wrong and the answer is E(Y) = E(X) + 1 (and you can calculate E(X) from f(x) integrated of course). Why is this?

Edit: I can visualize why the mean would be shifted when you translate by one unit in the x-direction. But why does the calculation not work?
If you recall your transformations of graphs acting on the x,y axes.

NOTE: x,y is not the same axes as X,Y random variables.

Then the linear combination of rv.s is acting on the y-axis, after the function has been evaluated, whereas here you have f(x-1), which is a change in the x-axis.

If you go back to first principles,

\displaystyle E(Y) = \int_{-\infty}^{\infty}yg(y)\; dy

\displaystyle E(Y) = \int_{-\infty}^{\infty}yf(y-1)\; dy

Let x = y-1

\displaystyle E(Y) = \int_{-\infty}^{\infty}(x+1)f(x)\; dx

\displaystyle \int_{-\infty}^{\infty}xf(x)\; dx +\int_{-\infty}^{\infty}f(x)\; dx

=E(X) + 1
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