Find the best upper bound on the probability that I catch at least 5 salmon given that the total number of fish I catch is 10 on average Let X be the number of fishes that I fish every day. Assume that $X \sim Poisson(\lambda)$. Each fish that I fish is a salmon with probability $0.3$, a tuna with probability $0.4$, a swordfish with probability $0.2$ and a carp with probability $0.1$.

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*Are the numbers of salmons, tunas, swordfishes and carps that I fish every day independent?

*Using some concentration inequality, give an upper bound on the probability that I fish at least 5 salmons given that the total number of fish that I get is on average 10 (you can admit that the number of salmons is distributed as $Poisson(\lambda p_1)$). Make sure that you obtain the best possible bound from the different concentration inequalities that you know.

My Attempt

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*For this one, Since you catch on average 10 fish a day, this would mean that if you catch more salmon, then you would catch fewer other types of fish, so intuitively, the number of salmons, tunas, swordfish, and carp are not independent. However, this is not formal enough.
Let $W$ denote the number of salmon per day, $V$ the number of tuna, $Y$ the number of swordfishes and $Z$ the number of carp. Then I to show independence, I need to show
$$\mathbb{P}(W=w)\mathbb{P}(V=v)=\mathbb{P}(W=w, V=v)$$ and so on for the other random variables. Would the random variables for each fish also be a Poisson distribution? If so, how would I find $\mathbb{P}(W=w, V=v)$?

*The concentration inequalities I know are Markov's inequality, Chebyshev's inequality, and Chernoff's bound. The formulas are
$$\mathbb{P}(X\geq c)\leq\frac{\mathbb{E}[X]}{c}$$
$$\mathbb{P}(|X-\mathbb{E}[X]|\geq c)\leq\frac{Var(X)}{c^2}$$
$$\mathbb{P}(X\geq c)\leq\frac{\mathbb{E}[e^{tX}]}{e^{ct}}$$
My question for this part is, if we are only given the average number of fish caught a day, how would we figure out the average number of salmons caught? Also, is it always true that Chebyshev's inequality is more accurate than Markov's and Chernoff's bound is the best?

 A: It is not necessary to use any kind of bounds. The average number of fish you catch per day is $10$, so $\lambda=10$. The distribution of the number of salmon you catch per day is then $S\sim\text{Poisson}(10\times .3)=\text{Poisson}(3)$. The probability that $S\ge5=1-(.0498+.1494+.2240+.2240+.1680)=.1848$.
For part a, we could try $P(S=s \&T=t)$ (standing for salmon and trout caught) equals $\sum_{f=0}^\infty P(S=s\& T=t|F=f)P(F=f)=\sum_{f=0}^\infty P(S=s|T=t,F=f)P(T=t|F=f)P(F=f)$
Trying $P(S=0,T=0)=^? P(S=0)P(T=0)$ we get the rhs to be $e^{-.3\lambda}e^{-.4\lambda}=e^{-.7\lambda}$.
The lhs is found by the total law of probability to be $\sum_{f=0}^\infty P(S=0|T=0,F=f)P(T=0|F=f)P(F=f)=\sum_{f=0}^\infty (.5)^f(.6)^f\frac{e^{-\lambda}\lambda^f}{f!}=e^{-\lambda}\sum_{f=0}^\infty \frac{(.3\lambda)^f}{f!}$
Using the identity $e^\lambda=\sum_{x=0}^\infty \frac{\lambda^x}{x!}$, we get the preceding expression to be $e^{-\lambda}e^{.3\lambda}=e^{-.7\lambda}$ which, surprisingly or not, matches the marginals multiplied together. Now I dunno if this will hold if $P(S=1)P(T=1)=^?P(S=1,T=1)$. Edit: Holds as well. Now would hazard to surmise that holds for all fishes and all quantities; conclude independence.
Proof for salmon and trout (get ready...):
$$\begin{split}P(S=s \&T=t)&=\sum_{f=s+t}^\infty P(S=s\& T=t|F=f)P(F=f)\\
&=\sum_{f=s+t}^\infty P(S=s|T=t,F=f)P(T=t|F=f)P(F=f)\\
&=\sum_{f=s+t}^\infty \frac{e^{-\lambda}\lambda ^f}{f!}{f\choose t}.4^t.6^{f-t}{f-t\choose s}.5^{f-t}\\
&=e^{-\lambda}\sum_{f=s+t}^\infty \frac{\lambda ^ff!t!(f-t)!s!(f-t-s)!}{f!.4^t.3^{f-t}(f-t)!}\\
&=\frac{(.3\lambda)^s\lambda ^t e^{-\lambda}.4^t}{t!s!}\sum_{f=s+t}^\infty \frac{(.3\lambda)^{f-t-s}}{(f-t-s)!}\\
&=\frac{(.3\lambda)^s\lambda ^t e^{-\lambda}.4^t}{t!s!}\sum_{x=0}^\infty \frac{(.3\lambda)^{x}}{x!}\\
&=\frac{(.3\lambda)^s}{s!}e^{-.3\lambda}\frac{e^{-.4\lambda}(.4\lambda)^t}{t!}=P(S=s)P(T=t)\end{split}$$
The Markov bound is $P(S\ge5)\le\frac{E(S)}{5}=\frac 35=.6$, which is true as we found above.
The Chernoff bound is $P(S\ge 5)\le\min_{s>0}e^{-5s}e^{3(e^s-1)}$. Since $\log$ is increasing this is the same as minimizing the log of it, $-5s+3e^s-3$. The derivative wrt $s$ is $-5+3e^s=0$, with solution at $s=\log {\frac 53}$. The second derivative is positive, indicating a minimum. This gives the bound $e^{-5\log{\frac 53}+5-3}=0.574573$.
