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x, y are independent possion variates. variance of X+Y = 9 P(X = 3/X+Y=6) = 5/54

Can anyone help me find the mean of X?

Does Chebyshev's inequality come into picture?

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1 Answer 1

I don't see what you want to do with Chebyshev's inequality.

Let $\lambda_X$ be the parameter for $X$ and $\lambda_Y$ be the parameter for Y.

Then $$9 = Var[X+Y] = Var[X] + Var[Y] = \lambda_X + \lambda_Y$$

and $$P(X = 3 | X+Y = 6) = P(Y = 3) = \frac{\lambda_Y^3 e^{-\lambda_Y}}{6} = \frac{5}{54}$$

This isn't a nice-looking system but you should be able to solve for the mean $\lambda_X$.

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